RD Sharma Solutions Class 10 Trigonometric Ratios Exercise 5.2

RD Sharma Solutions Class 10 Chapter 5 Exercise 5.2

RD Sharma Class 10 Solutions Chapter 5 Ex 5.2 PDF Free Download

Exercise: 5.2

 

Evaluate each of the following:

 Q 1 . sin \(45^{\circ}\) sin \(30^{\circ}\) + cos \(45^{\circ}\) cos \(30^{\circ}\)

 

Solution:

Sin \(45^{\circ}\)sin \(30^{\circ}\)  + cos \(45^{\circ}\) cos \(30^{\circ}\)                                                                                            [1]

We know that by trigonometric ratios we have ,

\(sin 45^{\circ}= \frac{1}{\sqrt{2}}\)                      \(sin 30^{\circ}= \frac{1}{2}\)

\(cos 45^{\circ}= \frac{1}{\sqrt{2}}\)                     \(cos 30^{\circ}= \frac{\sqrt{3}}{2}\)

Substituting the values in equation 1 , we get

\(\frac{1}{\sqrt{2}}\cdot \frac{1}{2}+\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2}\)

= \(\frac{1}{\sqrt{2}}\cdot \frac{\sqrt{3}}{2\sqrt{2}}\)

= \(\frac{\sqrt{3}+1}{2\sqrt{2}}\)

 

Q 2 . sin \(60^{\circ}\) cos \(30^{\circ}\) + cos \(60^{\circ}\) sin \(30^{\circ}\)

 

Solution:

sin \(60^{\circ}\) cos \(30^{\circ}\) + cos \(60^{\circ}\) sin \(30^{\circ}\)                                                                                             [1]

By trigonometric ratios we have ,

\(sin 60^{\circ}= \frac{\sqrt{3}}{2}\)                      \(sin 30^{\circ}= \frac{1}{2}\)

\(cos 30^{\circ}= \frac{\sqrt{3}}{2}\)                     \(cos 60^{\circ}= \frac{1}{2}\)

Substituting the values in equation 1 , we get

= \(\frac{\sqrt{3}}{2}\cdot \frac{\sqrt{3}}{2}+\frac{1}{2}\cdot \frac{1}{2}\)

= \(\frac{3}{4}+\frac{1}{4}\) = \(\frac{4}{4}\) = 1

 

Q 3 . cos \(60^{\circ}\) cos \(45^{\circ}\) – sin \(60^{\circ}\) sin \(45^{\circ}\)

 

Solution:

cos \(60^{\circ}\) cos \(45^{\circ}\) – sin \(60^{\circ}\) sin \(45^{\circ}\)                                                                                              [1]

We know that by trigonometric ratios we have ,

\(cos 60^{\circ}= \frac{1}{2}\)                                                        \(cos 45^{\circ}= \frac{1}{\sqrt{2}}\)

\(sin 60^{\circ}= \frac{\sqrt{3}}{2}\)                      \(sin 45^{\circ}= \frac{1}{\sqrt{2}}\)

Substituting the values in equation 1 , we get

\(\frac{1}{2}\cdot \frac{1}{\sqrt{2}}-\frac{\sqrt{3}}{2}\cdot \frac{1}{\sqrt{2}}\)

= \(\frac{1-\sqrt{3}}{2\sqrt{2}}\)

 

Q.4: \(sin^{2}30^{\circ}+sin^{2}45^{\circ}+sin^{2}60^{\circ}+sin^{2}90^{\circ}\)

 

Solution:

\(sin^{2}30^{\circ}+sin^{2}45^{\circ}+sin^{2}60^{\circ}+sin^{2}90^{\circ}\)                              [1]

We know that by trigonometric ratios we have ,

\(sin 30^{\circ}= \frac{1}{2}\)                    \(sin 45^{\circ}= \frac{1}{\sqrt{2}}\)

\(sin 60^{\circ}= \frac{\sqrt{3}}{2}\)   \(sin90^{\circ}\) = 1

Substituting the values in equation 1 , we get

= \(\left [ \frac{1}{2} \right ]^{2}+\left [ \frac{1}{\sqrt{2}} \right ]^{2}+\left [ \frac{\sqrt{3}}{2} \right ]^{2}+1\)

= \(\frac{1}{4}+\frac{1}{2}+\frac{3}{4}+1\)

= \(\frac{5}{2}\)

 

Q 5. \(cos^{2}30^{\circ}+cos^{2}45^{\circ}+cos^{2}60^{\circ}+cos^{2}90^{\circ}\)

 

Solution:

\(cos^{2}30^{\circ}+cos^{2}45^{\circ}+cos^{2}60^{\circ}+cos^{2}90^{\circ}\)              [1]

We know that by trigonometric ratios we have ,

\(cos 30^{\circ}= \frac{\sqrt{3}}{2}\)                     \(cos 45^{\circ}= \frac{1}{\sqrt{2}}\)

\(cos 60^{\circ}= \frac{1}{2}\)                                                        \(cos90^{\circ}\) = 0

Substituting the values in equation 1 , we get

\(\left [ \frac{\sqrt{3}}{2} \right ]^{2}+\left [ \frac{1}{\sqrt{2}} \right ]^{2}+\left [ \frac{1}{2} \right ]^{2}+0\)

= \(\frac{3}{4}+\frac{1}{2}+\frac{1}{4}\)

= \(\frac{3}{2}\)

 

Q 6 . \(tan^{2}30^{\circ}+tan^{2}45^{\circ}+tan^{2}60^{\circ}\)

 

Solution:

\(tan^{2}30^{\circ}+tan^{2}45^{\circ}+tan^{2}60^{\circ}\)                                               [1]

We know that by trigonometric ratios we have ,

\(tan30^{\circ}=\frac{1}{\sqrt{3}}\)                                    \(tan60^{\circ}=\sqrt{3}\)

\(tan45^{\circ}=1\)

Substituting the values in equation 1 , we get

\(\left [ \frac{1}{\sqrt{3}} \right ]^{2}+\left [ \sqrt{3} \right ]^{2}+1\)

= \(\frac{1}{3} +3+1\)

= \(\frac{13}{3}\)

 

Q 7 . \(2sin^{2}30^{\circ}-3cos^{2}45^{\circ}+tan^{2}60^{\circ}\)

 

Solution:

\(2sin^{2}30^{\circ}-3cos^{2}45^{\circ}+tan^{2}60^{\circ}\)                    [1]

We know that by trigonometric ratios we have ,

\(sin 30^{\circ}= \frac{1}{2}\)                                \(cos 45^{\circ}= \frac{1}{\sqrt{2}}\)

\(tan60^{\circ}=\sqrt{3}\)

Substituting the values in equation 1 , we get

= \(2\left ( \frac{1}{2} \right )^{2}-3\left ( \frac{1}{\sqrt{2}} \right )^{2}+\left ( \sqrt{3} \right )^{2}\)

= \(2\left ( \frac{1}{4} \right )-3\left ( \frac{1}{2} \right )+3\)

= \(\frac{1-3+6}{2}\)

= 2

 

Q8:\(sin^{2}30^{\circ}cos^{2}45^{\circ}+4tan^{2}30^{\circ}+\frac{1}{2}sin^{2}90^{\circ}-2cos^{2}90^{\circ}+\frac{1}{24}cos^{2}0^{\circ}\)

 

Solution:

\(sin^{2}30^{\circ}cos^{2}45^{\circ}+4tan^{2}30^{\circ}+\frac{1}{2}sin^{2}90^{\circ}-2cos^{2}90^{\circ}+\frac{1}{24}cos^{2}0^{\circ}\)                                          [1]

We know that by trigonometric ratios we have ,

\(sin 30^{\circ}= \frac{1}{2}\)

\(cos 45^{\circ}= \frac{1}{\sqrt{2}}\)

\(tan30^{\circ}=\frac{1}{\sqrt{3}}\)

\(sin90^{\circ}\) = 1

\(cos90^{\circ}\) = 0

\(cos0^{\circ}\) = 1

Substituting the values in equation 1 , we get

\(\left [ \frac{1}{2} \right ]^{2}\cdot \left [ \frac{1}{\sqrt{2}} \right ]^{2}+4\left [ \frac{1}{\sqrt{3}} \right ]^{2}+\frac{1}{2}\left [ 1 \right ]^{2}-2\left [ 0 \right ]^{2}+\frac{1}{24}\left [ 1 \right ]^{2}\)

= \(\frac{1}{8}+\frac{4}{3}+\frac{1}{2}+\frac{1}{24}\)

= \(\frac{48}{24}\) = 2

 

Q 9 . \(4\left ( sin^{4} 60^{\circ}+cos^{4} 30^{\circ}\right )-3\left ( tan^{2}60^{\circ}-tan^{2}45^{\circ} \right )+5cos^{2}45^{\circ}\)

 

Solution:

\(4\left ( sin^{4} 60^{\circ}+cos^{4} 30^{\circ}\right )-3\left ( tan^{2}60^{\circ}-tan^{2}45^{\circ} \right )+5cos^{2}45^{\circ}\)                                       [1]

We know that by trigonometric ratios we have ,

\(sin 60^{\circ}= \frac{\sqrt{3}}{2}\)                      \(cos 45^{\circ}= \frac{1}{\sqrt{2}}\)

\(tan60^{\circ}=\sqrt{3}\)                                        \(cos 30^{\circ}= \frac{\sqrt{3}}{2}\)

Substituting the values in equation 1 , we get

\(4\left ( \left [ \frac{\sqrt{3}}{2} \right ]^{4} +\left [ \frac{\sqrt{3}}{2} \right ]^{4} \right )-3 \left ( 3 \right )^{2}-1^{2}+5\left [ \frac{1}{\sqrt{2}} \right ]^{2}\)

= \(4\cdot \frac{18}{16}-6+\frac{5}{2}\)

= \(\frac{1}{4}-6+\frac{5}{2}\)

= \(\frac{14}{2}-6\) = 7 – 6 = 1

 

Q 10 . \(\left (cosec^{2}45^{\circ}sec^{2}30^{\circ} \right )\left ( sin^{2}30^{\circ} +4cot^{2}45^{\circ}-sec^{2}60^{\circ}\right )\)

 

Solution:

\(\left (cosec^{2}45^{\circ}sec^{2}30^{\circ} \right )\left ( sin^{2}30^{\circ} +4cot^{2}45^{\circ}-sec^{2}60^{\circ}\right )\)                                                                        [1]

We know that by trigonometric ratios we have ,

\(cosec45^\circ=\sqrt{2}\)                            \(sec30^\circ=\frac{2}{\sqrt{3}}\)

\(sin 30^{\circ}= \frac{1}{2}\)                    \(cot45^{\circ}\) = 1

\(sec60^{\circ}\) = 2

Substituting the values in equation 1 , we get

\(\left ( \left [ \sqrt{2} \right ]^{2}.\left [ \frac{2}{\sqrt{3}} \right ]^{2} \right )\left ( \left [ \frac{1}{2} \right ]^{2} +4\left ( 1 \right )\left ( 2 \right )^{2}\right )\)

= \(3\cdot \frac{4}{3}\cdot \frac{1}{4}\)

= \(\frac{2}{3}\)

 

Q11. \(cosec^{3}30^{\circ} cos 60^{\circ} tan^{3}45^{\circ} sin^{2}90^{\circ} sec^{2}45^{\circ} cot 30^{\circ}\)

 

Solution:

Given,

= \(cosec^{3}30^{\circ} cos 60^{\circ} tan^{3}45^{\circ} sin^{2}90^{\circ} sec^{2}45^{\circ} cot 30^{\circ}\)

= \(2^{3} (\frac{1}{2}) (1^{3}) (1^{2}) (\sqrt{2}^{2}) (\sqrt{3})\)

= \((2)^{3}\times (\frac{1}{2})\times (1^{3})\times (1^{2})\times (\sqrt{2}^{2}) \times (\sqrt{3})\)

= \(8\times (\frac{1}{2})\times (1)\times (1)\times (2) \times (\sqrt{3})\)

= \(8\sqrt{3}\)

 

Q12. \(cot^{2}30^{\circ}- 2cos^{2}60^{\circ}-\frac{3}{4}sec^{2}45^{\circ} – 4sec^{2}30^{\circ}\)

Solution:

Given,

= \(cot^{2}30^{\circ}- 2cos^{2}60^{\circ}-\frac{3}{4}sec^{2}45^{\circ} – 4sec^{2}30^{\circ}\)

= \((\sqrt{3}^{2})\times 2(\frac{1}{2})^{2}\times (\frac{3}{4}\times \sqrt{2}^{2})\times (4\times (\frac{2}{\sqrt{3}})^{2})\)

= \(3-\frac{1}{2}-\frac{3}{2}-\frac{16}{3}\)

= \(\frac{-13}{3}\)

 

Q13. \((cos 0^{\circ}+sin 45^{\circ}+sin 30^{\circ})(sin 90^{\circ}-cos 45^{\circ}+cos 60^{\circ})\)

 

Solution:

Given,

\((cos 0^{\circ}+sin 45^{\circ}+sin 30^{\circ})(sin 90^{\circ}-cos 45^{\circ}+cos 60^{\circ})\\ (1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}})(1 – \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}})\\ (\frac{3}{2}+\frac{1}{\sqrt{2}})(\frac{3}{2}-\frac{1}{\sqrt{2}})\\ ((\frac{3}{2})^{2}-(\frac{1}{\sqrt{2}})^{2}) \frac{9}{4}-\frac{1}{2} \frac{7}{4}\)

 

Q14. \(\frac{sin 30^{\circ}-sin 90^{\circ} + 2 cos 0^{\circ}}{tan 30^{\circ} tan 60^{\circ}}\)

Solution:

Given,

\(\frac{sin 30^{\circ}-sin 90^{\circ} + 2 cos 0^{\circ}}{tan 30^{\circ} tan 60^{\circ}}\\ \frac{\frac{1}{2}-1+2}{\frac{1}{\sqrt{3}}\times\sqrt{3}}\\ \frac{3}{2}\)

 

Q15. \(\frac{4}{cot^{2}30^{\circ}}+\frac{1}{sin^{2}60^{\circ}}-cos^{2}45^{\circ}\)

 

Solution:

Given,

\(\frac{4}{cot^{2}30^{\circ}}+\frac{1}{sin^{2}60^{\circ}}-cos^{2}45^{\circ}\\ = \frac{4}{(\sqrt{3})^{2}} + \frac{1}{(\frac{\sqrt{3}}{2})^{2}} – (\frac{1}{\sqrt{2}})^{2}\\ = \frac{4}{3} + \frac{4}{3} – \frac{1}{2}\\ = \frac{16 – 3}{6}\\ = \frac{13}{6}\)

 

Q16. \(4(sin^{4}30^{\circ}+cos^{2}60^{\circ})-3(cos^{2}45^{\circ}-sin^{2}90^{\circ})-sin^{2}60^{\circ}\)

Solution:

Given,

\(4(sin^{4}30^{\circ}+cos^{2}60^{\circ})-3(cos^{2}45^{\circ}-sin^{2}90^{\circ})-sin^{2}60^{\circ}\\ = 4((\frac{1}{2})^{4}+(\frac{1}{2})^{2})-3((\frac{1}{\sqrt{2}})^{2}-1)-(\frac{\sqrt{3}}{2})^{2}\\ = 4(\frac{1}{16}+\frac{1}{4})+\frac{3}{2}-\frac{3}{4}\\ = \frac{8}{4} = 2\)

 

Q17. \(\frac{tan^{2}60^{\circ} + 4cos^{2}45^{\circ} + 3sec^{2}30^{\circ} + 5cos^{2}90^{\circ}}{cosec 30^{\circ}+ sec 60^{\circ}-cot^{2}30^{\circ}}\)

 

Solution:

Given,

\(\frac{tan^{2}60^{\circ} + 4cos^{2}45^{\circ} + 3sec^{2}30^{\circ} + 5cos^{2}90^{\circ}}{cosec 30^{\circ}+ sec 60^{\circ}-cot^{2}30^{\circ}}\\ = \frac{(\sqrt{3})^{2}+4(\frac{1}{\sqrt{2}})^{2}+3(\frac{2}{\sqrt{3}})^{2}+5(0)}{2 + 2 – (\sqrt{3})^{2}}\\ = 3 + 2 + 4\\ = 9\)

 

Q18. \(\frac{sin 30^{\circ}}{sin 45^{\circ}} + \frac{tan 45^{\circ}}{sec 60^{\circ}} – \frac{sin 60^{\circ}}{cot 45^{\circ}} – \frac{cos 30^{\circ}}{sin 90^{\circ}}\)

 

Solution:

Given,

\(\frac{sin 30^{\circ}}{sin 45^{\circ}} + \frac{tan 45^{\circ}}{sec 60^{\circ}} – \frac{sin 60^{\circ}}{cot 45^{\circ}} – \frac{cos 30^{\circ}}{sin 90^{\circ}}\\ = \frac{\frac{1}{2}}{\frac{1}{\sqrt{2}}} + \frac{1}{2} – \frac{\frac{\sqrt{3}}{2}}{1} – \frac{\frac{\sqrt{3}}{2}}{1}\\ = \frac{\sqrt{2}}{2} + \frac{1}{2} – \frac{\sqrt{3}}{2} – \frac{\sqrt{3}}{2}\\ = \frac{\sqrt{2} + 1 – 2\sqrt{3}}{2}\)

 

Q19. \(\frac{tan 45^{\circ}}{cosec 30^{\circ}} + \frac{sec 60^{\circ}}{cot 45^{\circ}} + \frac{s sin 90^{\circ}}{2 cos 0^{\circ}}\)

 

Solution:

Given,

\(\frac{tan 45^{\circ}}{cosec 30^{\circ}} + \frac{sec 60^{\circ}}{cot 45^{\circ}} + \frac{s sin 90^{\circ}}{2 cos 0^{\circ}}\\ = \frac{1}{2} + \frac{2}{1} – \frac{5(1)}{2(1)}\\ = \frac{5}{2} – \frac{5}{2}\\ = 0\)

 

Q20. \(2 sin 3x = \sqrt{3}\)

 

Solution:

Given,

\(2 sin 3x = \sqrt{3}\\ => sin 3x = \frac{\sqrt{3}}{2}\\ => sin 3x = sin 60^{\circ}\\ => 3x = 60^{\circ}\\ => x = 20^{\circ}\)

 

Q21) \(2sin \frac{x}{2}=1,\;x=?\)

Solution:

\(sin \frac{x}{2}=\frac{1}{2}\)

\(sin \frac{x}{2}=sin30^{0}\)

\(\frac{x}{2}=30^{0}\)

x = 600

 

Q22) \(\sqrt{3}sin x=cos x\)

 

Solution:

\(\sqrt{3}\;tan x=1\)

\(tan x=\frac{1}{\sqrt{3}}\)

\(∴ tanx=tan45^{0}\)

x = 450

 

Q23) Tan x = sin 450 cos 450 + sin 300

 

Solution:

\(Tanx=\frac{1}{\sqrt{2}}.\frac{1}{\sqrt{2}}+\frac{1}{2}\;\;\;\;\;[∵ sin 45^{0}=\frac{1}{\sqrt{2}}\;cos45^{0}=\frac{1}{\sqrt{2}}\;sin30^{0}=\frac{1}{2}]\)

\(Tanx=\frac{1}{2}+\frac{1}{2}\)

Tan x = 1

Tan x = 450

x = 450

 

Q24) \(\sqrt{3}\;Tan2x=cos60^{0}+sin45^{0}cos45^{0}\)

 

Solution:

\(\sqrt{3}\;Tan2x=\frac{1}{2}+\frac{1}{\sqrt{2}}.\frac{1}{\sqrt{2}}\;\;\;\;\;\;[∵ cos60^{0}=\frac{1}{2}\;sin45^{0}=cos45^{0}=\frac{1}{\sqrt{2}}]\)

\(\sqrt{3}\;Tan2x=\frac{1}{\sqrt{3}}\Rightarrow tan2x=tan30^{0}\)

2x = 300

x = 150

 

Q25) \(cos2x=cos60^{0}\;cos30^{0}+sin60^{0}sin30^{0}\)

 

Solution:

\(cos2x=\frac{1}{2}.\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}.\frac{1}{2}\;\;\;\;\;\;[∵ cos60^{0}=sin30^{0}=\frac{1}{2}sin60^{0}=cos30^{0}=\frac{\sqrt{3}}{2}]\)

\(cos2x=2.\frac{\sqrt{3}}{4}\)

\(cos2x=\frac{\sqrt{3}}{2}\)

\(cos2x=cos30^{0}\)

\(2x=30^{0}\)

\(x=15^{0}\)

 

Q26) \(If\;\theta=30^{0},\;verify\\\\ (i)Tan2\theta=\frac{2Tan\theta}{1-tan^{2}\theta}\)

 

Solution:

\(Tan2\theta=\frac{2Tan\theta}{1-tan^{2}\theta}…..(i)\)

Substitute \(\theta=30^{0}\) in equation (i)

LHS = Tan 600 = \(\sqrt{3}\)

RHS = \(\frac{2Tan 30^{0}}{1+(Tan 30^{0})^{2}}=\frac{2-\frac{1}{\sqrt{2}}}{1-(\frac{1}{\sqrt{2}})^{2}}=\sqrt{3}\)

Therefore, LHS = RHS

(ii) \(sin\theta=\frac{2tan\theta}{1-tan^{2}\theta}\)

Substitute \(\theta=30^{0}\)

\(sin60^{0}=\frac{2tan30^{0}}{(1-tan30^{0})^{2}}\)

=>\(\frac{\sqrt{3}}{2}=\frac{2.\frac{1}{\sqrt{2}}}{1+(\frac{1}{\sqrt{2}})^{2}}\)

=> \(\frac{\sqrt{3}}{2}=\frac{2}{\sqrt{3}}.\frac{3}{4}\\ \Rightarrow \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2}\)

Therefore, LHS = RHS.

 

(iii) \(cos2\theta=\frac{1-tan^{2}\theta}{1+tan^{2}\theta}\)

Substitute \(\theta=30^{0}\)

LHS = \(cosec\theta\)                       RHS = \(\frac{1-tan^{2}\theta}{1+tan^{2}\theta}\)

= cos 2(300)                                                    \(=\frac{1-tan^{2}30^{0}}{1+tan^{2}30^{0}}\)

Cos 600 = \(\frac{1}{2}\)                 \(= \frac{1-(\frac{1}{\sqrt{2}})^{2}}{1+(\frac{1}{\sqrt{2}})^{2}}=\frac{\frac{2}{2}}{\frac{1}{2}}=\frac{1}{2}\)

Therefore, LHS = RHS

 

(iv) \(cos3\theta=4cos^{3}\theta-3cos\theta\)

Solution:

LHS = \(cos3\theta\)

Substitute \(\theta=30^{0}\)

= cos 3 (300) = cos 900

= 0

RHS = \(4cos^{3}\theta-3cos\theta\)

= \(4cos^{3}30^{0}-3cos30^{0}\)

= \(4(\frac{\sqrt{3}}{2})^{3}-3.\frac{\sqrt{3}}{2}\)

= \(3.\frac{\sqrt{3}}{2}-3.\frac{\sqrt{3}}{2}\)

= 0

Therefore, LHS = RHS.

 

Q27) If A = B = 600. Verify  (i) Cos (A – B) = Cos A Cos B + Sin A Sin B

 

Solution:

Cos (A – B) = Cos A Cos B + Sin A Sin B…….(i)

Substitute A and B in (i)

=>cos (600 – 600) = cos 600 cos 600 + sin 600 sin 600

=>cos 00 = \((\frac{1}{2})^{2}+(\frac{\sqrt{3}}{2})^{2}\)

=>1 = \(\frac{1}{4}+\frac{3}{4}\)

=>1 = 1

Therefore, LHS = RHS

 

(ii) Substitute A and B in (i)

=>sin (600 – 600) = sin 600 cos 600 – cos 600 sin 600

=> sin 00 = 0

=>0 = 0

Therefore, LHS = RHS

 

(iii) \(Tan (A-B)=\frac{TanA-TanB}{1+TanA\;TanB}\)

A = 600, B = 600 we get,

\(Tan (60^{0}-60^{0})=\frac{Tan60^{0}-Tan60^{0}}{1+Tan60^{0}\;Tan60^{0}}\)

Tan 00 = 0

0 = 0

Therefore, LHS = RHS

 

Q28 ) If A = 300, B = 600 verify:

(i) Sin (A + B) = Sin A Cos B + Cos A Sin B

 

Solution:

A = 300, B = 600 we get

Sin (300 + 600) = Sin 300  Cos 600  + Cos 300  Sin 600

Sin (900) = \(\frac{1}{2}.\frac{1}{2}+\frac{\sqrt{3}}{2}.\frac{\sqrt{3}}{2}\)

Sin (900) = 1 => 1 = 1

Therefore, LHS = RHS

 

(ii) Cos (A + B) = Cos A Cos B – Sin A Sin B

A = 300, B = 600 we get

Cos (300 + 600) = Cos 300 Cos 600 – Sin 300 Sin 600

Cos (900) = \(\frac{1}{2}.\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}.\frac{1}{2}\)

0 = 0

Therefore, LHS = RHS

Q29. If sin(A+B) = 1 and cos(A-B) = 1, \(0^{\circ}< A+B\leq 90^{\circ}\), A≥B find A and B.

 

Sol:

Given,

sin(A+B) = 1 this can be written as sin(A+B) = \(sin(90^{\circ})\)

cos(A-B) = 1 this can be written as cos(A-B) = \(cos(0^{\circ})\)

=> A + B = \(90^{\circ}\)

A – B = \(0^{\circ}\)

2A       = \(90^{\circ}\)

A         = \(\frac{90^{\circ}}{2}\)

A         = \(45^{\circ}\)

Substitute A value in A – B = \(0^{\circ}\)

\(45^{\circ}\) – B = \(0^{\circ}\)

B =\(45^{\circ}\)

Hence, the value of A = \(45^{\circ}\) and B =\(45^{\circ}\)

 

Q30. If tan(A-B) = \(\frac{1}{\sqrt{3}}\) and tan(A+B) = \(\sqrt{3}\), \(0^{\circ}< A+B\leq 90^{\circ}\), A>B find A and B

Solution:

Given,

tan(A-B) = \(\frac{1}{\sqrt{3}}\)

A – B   = \(tan^{-1}(\frac{1}{\sqrt{3}})\)

A – B   = \(30^{\circ}\)                  ——- 1

tan(A+B) = \(\sqrt{3}\)

A + B = \(tan^{-1}{\sqrt{3}}\)

A + B  = \(60^{\circ}\)                   ——- 2

Solve equations 1 and 2

A + B = \(30^{\circ}\)

A – B = \(60^{\circ}\)

2A  = \(90^{\circ}\)

A = \(\frac{90^{\circ}}{2}\)

A = \(45^{\circ}\)

Substitute the value of A in equation 1

\(45^{\circ}\) + B =  \(30^{\circ}\)

B = \(30^{\circ}\)\(45^{\circ}\)

B = \(15^{\circ}\)

The value of A = \(45^{\circ}\) and B = \(15^{\circ}\)

 

Q31. If sin(A-B) = \(\frac{1}{2}\) and cos(A+B) = \(\frac{1}{2}\), \(0^{\circ}< A+B\leq 90^{\circ}\), A<B find A and B.

 

Solution:

Given,

sin(A-B) = \(\frac{1}{2}\)

A – B = \(sin^{-1}(\frac{1}{2})\)

A – B = \(30^{\circ}\)                    ——- 1

cos(A+B) = \(\frac{1}{2}\)

A + B = \(cos^{-1}(\frac{1}{2})\)

A + B = \(60^{\circ}\)                 ——- 2

Solve equations 1 and 2

A + B = \(60^{\circ}\)

A – B = \(30^{\circ}\)

2A  = \(90^{\circ}\)

A = \(\frac{90^{\circ}}{2}\)

A = \(45^{\circ}\)

Substitute the value of A in equation 2

\(45^{\circ}\) + B =  \(60^{\circ}\)

B = \(60^{\circ}\)\(45^{\circ}\)

B = \(15^{\circ}\)

The value of A = \(45^{\circ}\) and B = \(15^{\circ}\)

 

Q32. In a \(\Delta\) ABC right angled triangle at B, \(\angle A\;=\;\angle C\).Find the values of:

1. sinAcosC + cosAsinC

 

Solution:

since, it is given as \(\angle A\;=\;\angle C\)

the value of A and C is \(45^{\circ}\), the value of angle  B is \(90^{\circ}\)

because the sum of angles of triangle is \(180^{\circ}\)

=> sin(\(45^{\circ}\))cos(\(45^{\circ}\)) + cos(\(45^{\circ}\))sin(\(45^{\circ}\))

=> \((\frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}})\) + \((\frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}})\)

=> \(\frac{1}{2}\) + \(\frac{1}{2}\)

=> 1

The value of  sinAcosC + cosAsinC is 1

 

2. sinAsinB + cosAcosB

Solution:

since, it is given as \(\angle A\;=\;\angle C\)

the value of A and C is \(45^{\circ}\), the value of angle  B is \(90^{\circ}\)

because the sum of angles of triangle is \(180^{\circ}\)

=> sin(\(45^{\circ}\))sin(\(90^{\circ}\)) + cos(\(45^{\circ}\))sin(\(90^{\circ}\))

=> \(\frac{1}{\sqrt{2}}\)(1) + \(\frac{1}{\sqrt{2}}\)(0)

=> \(\frac{1}{\sqrt{2}}\) + 0

=> \(\frac{1}{\sqrt{2}}\)

The value of sinAsinB + cosAcosB is \(\frac{1}{\sqrt{2}}\)

 

Q33. Find the acute angle A and B, if sin(A+2B) = \(\frac{\sqrt{3}}{2}\) and cos(A+4B) = 0, A>B.

 

Solution:

Given,

sin(A+2B) = \(\frac{\sqrt{3}}{2}\)

A + 2B = \(sin^{-1}\frac{\sqrt{3}}{2}\)

A + 2B = \(60^{\circ}\)                     ——– 1

Cos(A+4B) = 0

A + 4B = \(sin^{-1}(90)\)

A + 4B = \(90^{\circ}\)                     ——– 2

Solve equations 1 and 2

A + 2B = \(60^{\circ}\)

A + 4B = \(90^{\circ}\)

(-)  (-)         (-)

-2B = –\(30^{\circ}\)

2B = \(30^{\circ}\)

B = \(\frac{30^{\circ}}{2}\)

B = \(15^{\circ}\)

Substitute B value in eq 2

A + 4B = \(90^{\circ}\)

A + 4(\(15^{\circ}\)) = \(90^{\circ}\)

A + \(60^{\circ}\) = \(90^{\circ}\)

A = \(90^{\circ}\)\(60^{\circ}\)

A = \(30^{\circ}\)

The value of A = \(30^{\circ}\) and B = \(15^{\circ}\)

 

Q 34. In \(\Delta PQR\), right angled at Q, PQ = 3 cm and PR = 6 cm. Determine \(\angle\) P and \(\angle\) R.

 

Solution:

Given,

In \(\Delta PQR\), right angled at Q, PQ = 3 cm and PR = 6 cm

By Pythagoras theorem,

\(PR^{2} = PQ^{2} + QR^{2}\\ => 6^{2} = 3^{2} + QR^{2}\\ => QR^{2} = 36 – 9\\ => QR = \sqrt{27}\\ => QR = 3\sqrt{3}\)

sin R =  \(\frac{3}{6} = \frac{1}{2} = sin 30^{\circ}\)

\(\angle R = 30^{\circ}\)

As we know, Sum of angles in a triangle = 180

\(\angle P + \angle Q + \angle R = 180^{\circ} \\ => \angle P + 90^{\circ} + 30^{\circ} = 180^{\circ}\\ => \angle P = 180^{\circ} – 120^{\circ}\\ => \angle P = 60^{\circ}\)

Therefore, \(\angle R = 30^{\circ}\)

And, \(\angle P = 60^{\circ}\)

 

Q35. If sin(A – B) = sin A cos B – cos A sin B and cos (A – B) = cos A cos B + sin A sin B, find the values of sin 15 and cos 15.

 

Solution:

Given,

sin(A – B) = sin A cos B – cos A sin B

And, cos (A – B) = cos A cos B + sin A sin B

We need to find, sin 15 and cos 15.

Let A = 45 and B = 30

sin 15 = sin (45- 30) = sin 45 cos 30 – cos 45 sin 30

= \(= (\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2}) – (\frac{1}{\sqrt{2}}\times \frac{1}{2})\\ = \frac{\sqrt{3}-1}{2\sqrt{2}}\)

cos 15 = cos (45- 30) = cos 45 cos 30 – sin 45 sin 30

= \(= (\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2}) + (\frac{1}{\sqrt{2}}\times \frac{1}{2})\\ = \frac{\sqrt{3}+1}{2\sqrt{2}}\)

 

Q36. In a right triangle ABC, right angled at C, if \(\angle B=60^{\circ}\) and AB=15 units. Find the remaining angles and sides. 

    25                                           

 

Solution:

\(sin 60^{\circ}=\frac{x}{15}\\ \frac{\sqrt{3}}{2}=\frac{x}{15}\\ x=\frac{15\sqrt{3}}{2}units\\ \\ \\ cos 60^{\circ}=\frac{x}{15}\\ \frac{1}{2}=\frac{x}{15}\\ x=\frac{15}{2}\\ x=7.5 units\)

 

 Q37. In \(\Delta\)ABC is a right triangle such that \(\angle C = 90^{\circ}\), \(\angle  A= 45^{\circ}\) and BC = 7 units. Find the remaining angles and sides.

 

Solution:

 

26

 

 

Here, \(\angle C = 90^{\circ}\) and \(\angle  A= 45^{\circ}\)

We know that,

\(\angle A +\angle B +\angle C\) = \(180^{\circ}\)

=> \(45^{\circ}\) + \(90^{\circ}\) + \(\angle C\) = \(180^{\circ}\)

=>  \(135^{\circ}\) + \(\angle C\) = \(180^{\circ}\)

=> \(\angle C\) = \(180^{\circ}\)\(135^{\circ}\)

=> \(\angle C\) = \(45^{\circ}\)

The value of the remaining angle C is \(45^{\circ}\)

Now, we need to find the sides x and y

here,

cos(45) = \(\frac{BC}{AB}\)

\(\frac{1}{\sqrt{2}}\) = \(\frac{7}{y}\)

y = \(7\sqrt{2}\) units

sin(45) = \(\frac{AC}{AB}\)

\(\frac{1}{\sqrt{2}}\) = \(\frac{x}{y}\)

\(\frac{1}{\sqrt{2}}\) = \(\frac{x}{7\sqrt{2}}\)

x = \(\frac{7\sqrt{2}}{\sqrt{2}}\)

x = 7 units

the value of x = 7 units and y = \(7\sqrt{2}\) units

 

Q 38 . In a rectangle ABCD , AB = 20 cm , \(\angle\)BAC = \(60^{\circ}\) , calculate side BC and diagonals AC and BD .

Solution:

Let AC = x cm and CB = y cm

Since , \(cos\theta\) = \(\frac{base}{hypotenuse}\)

Therefore , \(cos 60^{\circ}=\frac{20}{x}\)

\(\Rightarrow \frac{1}{2}=\frac{20}{x}\)                           [since,\(cos60^{\circ}=\frac{1}{2}\)]

\(\Rightarrow\) x= 40 cm = AC

Similarly BD = 40 cm

Now ,

Since , \(sin\theta\) = \(\frac{perpendicular}{hypotenuse}\)

Therefore , \(sin 60^{\circ}=\frac{BC}{AC}\)

\(\Rightarrow \frac{\sqrt{3}}2{}=\frac{y}{40}\)

\(\Rightarrow y=\frac{40\sqrt{3}}{2}\)

\(\Rightarrow y=20\sqrt{3}\) cm .

 

Q39:If A & B are acute angles such that tanA=1/2 tanB=1/3 and tan(A+B)= \(\frac{tanA+tanB}{1-tanA\;tanB}\), find A+B.

 

Solution:

\(Tan(A+B)=\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}.\frac{1}{3}}\)

\(=\frac{\frac{3+2}{6}}{\frac{5}{6}}\)

\(=\frac{\frac{5}{6}}{\frac{5}{6}}\)

\(Tan (A+B)=\frac{5}{6}\times \frac{6}{5}\)

\((A+B)=Tan^{-1}(1)\)

(A+B) = 450

 

Q 40: Prove that : \((\sqrt{3}-1)(3-cot30{^{\circ}})=tan^{3}60-2sin60^{\circ}\)

Ans:

L.H.S => \((\sqrt{3}+1)(3-cot30^{\circ})\)

=\((\sqrt{3}+1)(3-\sqrt{3})\;\;∵ cot 30^{\circ}=\sqrt{3}\)

=\((\sqrt{3}+1)(\sqrt{3}-1)\sqrt{3}\)

= \(((\sqrt{3})^{2}-(1)^{2})\sqrt{3}\)

= \(2\sqrt{3}\)

R.H.S => \(tan^{3}60-2sin60^{\circ}\)

= \((\sqrt{3})^{3}-2\times\frac{\sqrt{3}}{2}\)

= \(3\sqrt{3}-\sqrt{3}\)

= \(2\sqrt{3}\)

L.H.S = R.H.S

Hence proved.