RS Aggarwal Solutions Class 10 Trigonometric Ratios

RS Aggarwal Solutions Class 10 Chapter 5

 

Q1:

If \(sin\Theta =\frac{a}{b},\:show\:that\:\left ( sec\Theta +tan\Theta \right )=\sqrt{\frac{b+a}{b-a}}\)

Sol:

L.H.S = \(\left (sec\Theta +tan\Theta \right )\)

\(=\frac{1}{cos\Theta }+\frac{sin\Theta }{cos\Theta }\)
\(=\frac{1+sin\Theta }{cos\Theta }\)
\(=\frac{1+sin\Theta }{\sqrt{1-sin^{2}\Theta }}\)
\(=\frac{\left ( 1+\frac{a}{b} \right )}{\sqrt{1-\left ( \frac{a}{b} \right )^{2}}}\)
\(=\frac{\left ( \frac{1}{1}+\frac{a}{b} \right )}{\sqrt{\frac{1}{1}-\frac{a^{2}}{b^{2}}}}\)
\(=\frac{\left ( \frac{b+a}{b} \right )}{\sqrt{\frac{b^{2}-a^{2}}{b}}}\)
\(=\frac{\left ( \frac{b+a}{b} \right )}{\left ( \frac{\sqrt{b^{2}-a^{2}}}{b} \right )}\)
\(=\frac{\left ( b+a \right )}{\sqrt{\left ( b+a \right )\left ( b-a \right )}}\)
\(=\frac{\left ( b+a \right )}{\sqrt{\left ( b+a \right )}\sqrt{\left ( b-a \right )}}\)
\(=\frac{\sqrt{\left ( b+a \right )}}{\sqrt{\left ( b-a \right )}}\)
\(=\sqrt{\frac{b+a}{b-a}}\)

R.H.S

Q2:

If \(cos\Theta =\frac{3}{5},\:show\:that\:\frac{\left ( sin\Theta -cot\Theta \right )}{2tan\Theta }=\frac{3}{160}\)

Sol:

L.H.S = \(\frac{\left ( sin\Theta -cot\Theta \right )}{2tan\Theta }\)

\(=\frac{\left ( sin\Theta -\frac{cos\Theta }{sin\Theta } \right )}{2\left ( \frac{sin\Theta }{cos\Theta } \right )}\)
\(=\frac{\left ( \frac{sin^{2}\Theta -cos\Theta }{sin\Theta } \right )}{\left ( \frac{2sin\Theta }{cos\Theta } \right )}\)
\(=\frac{cos\Theta \left ( sin^{2}\Theta -cos\Theta \right )}{2sin^{2}\Theta }\)
\(=\frac{cos\Theta \left ( 1-cos^{2}\Theta -cos\Theta \right )}{2\left ( 1-cos^{2}\Theta \right )}\)
\(=\frac{\frac{3}{5}\left [ 1-\left ( \frac{3}{5} \right )^{2}-\frac{3}{5} \right ]}{2\left [ 1-\left ( \frac{3}{5} \right ) ^{2}\right ]}\)
\(=\frac{\frac{3}{5}\left ( \frac{1}{1}-\frac{9}{25}-\frac{3}{5} \right )}{2\left ( 1-\frac{9}{25} \right )}\)
\(=\frac{\frac{3}{5}\left ( \frac{25-9-15}{25} \right )}{2\left ( \frac{25-9}{25} \right )}\)
\(=\frac{\frac{3}{5}\left ( \frac{1}{25} \right )}{2\left ( \frac{16}{25} \right )}\)
\(=\frac{3}{5\times 2\times 16}\)
\(=\frac{3}{160}\)

R.H.S

Q3:

If \(\tan \theta = \frac{4}{3},\) show that \(\sin \theta +\cos \theta =\frac{7}{5}\)

Sol:

Given:

\(\tan \theta =\frac{BC}{AB}= \frac{4}{3}\)

Let AB = 4k and AC = 3k,

Where k is positive

Let us draw a \(\bigtriangleup ABC \) in which \(\angle B=90^{\circ}\; and \; \angle BAC=\theta\)

By pythagoras theorem, we have

\(AC^{2}=AB^{2}+BC^{2}\)
\(=(3k)^{2}+(4)k^{2}\)
\(=(9k^{2})+(16k^{2})\)

= \(=25k^{2}\)

AC = 5k

Therefore, \(\sin \theta= \frac{BC}{AC} = \frac{4k} {5k} = \frac{4} {5} \)

\( \cos \theta= \frac{AB} {AC} = \frac{3k} {5k} = \frac{3} {5} \)

Therefore, \(\sin \theta +\cos \theta =\left ( \frac{4}{5}+\frac{3}{5} \right )\)

\(Hence\; \; \sin \theta +\cos \theta =\frac{7}{5}\)

Q4:

If \( \tan \theta = \frac{a}{b} \), show that \( \frac{a\sin \theta – b \cos \theta } {a\sin \theta + b \cos \theta} = \frac{a^{2}- b^{2}}{a^{2}+b^{2}}\)

Sol:

Given:

\( \tan \theta = \frac{a}{b}= \frac{ak}{bk}= \frac{BC}{AB} \)

Let us draw a \(\bigtriangleup ABC \) in which \(\angle B = 90^{\circ}\; and \; \angle A=\theta\)

Pythagoras theorem, we have

\(AC^{2} = AB^{2} + BC^{2} = ??b^{2}k^{2} + a^{2}k^{2} \)

Therefore, \(AC = k \sqrt{a^{2}+b^{2}}\)

\(\sin \theta = \frac{BC}{AC} = \frac{ak}{k\sqrt{a^{2}+b^{2}}} = \frac{a}{\sqrt{a^{2}+b^{2}}}\)
\(\cos \theta = \frac{AB}{AC} = \frac{bk}{k\sqrt{a^{2}+b^{2}}} = \frac{b}{\sqrt{a^{2}+b^{2}}}\)

L.H.S. = \(\frac{a\sin \theta – b \cos \theta }{a\sin \theta + b \cos \theta}\)

\(= \frac{a \frac{a}{\sqrt{a^{2} + b^{2}}} – b. \frac{b}{\sqrt{a^{2} + b^{2}}}}{a.b} = \frac{\frac{a^{2}-b^{2}}{\sqrt{a^{2}+b^{2}}}}{\frac{a^{2} + b^{2}}{\sqrt{a^{2}+b^{2}}}} = \frac{a^{2}- b^{2}}{a^{2}+b^{2}}\)

= R.H.S.

Q5:

If \(3\tan \theta =4\), show that \(\frac{4\cos \theta – \sin \theta }{2\cos \theta + \sin \theta } =\frac{4}{5}\).

Sol:

Given:

\(3\tan \theta =4\)
\(\tan \theta = \frac{4}{3}\)

Let us draw a \(\bigtriangleup ABC \) in which \(\angle B = 90^{\circ}\; and \; \angle BAC=\theta\)

By pythagoras theorem, we have

\(AC^{2} = AB^{2} + BC^{2} \)
\( = (4k)^{2} + (3k)^{2}\)
\(= 16k^{2} + 9k^{2} = 25k^{2}\)

\( AC = 5k \),

Now,

\( \sin \theta= \frac{BC}{AB} = \frac{4k} {5k} = \frac{4} {5} \)
\( \cos \theta = \frac{AB}{AC}= \frac{3k} {5k} = \frac{3}{5} \)

L.H.S. = \(\frac{4\cos \theta – \sin \theta }{2\cos \theta + \sin \theta } = \frac{4\times \frac{\frac{3}{5}- \frac{4}{5}}{}}{2\times \frac{3}{4}+\frac{4}{5}}\)

\(\frac{12-4}{6+4}= \frac{8}{10}= \frac{4}{5}\) = R.H.S.

Q6:

If \(3\cot \theta = 2\), ??show that \(\frac{4\sin \theta – 3\cos \theta }{2\sin \theta +6\cos \theta }=\frac{1}{3}\)

Sol:

Given:

\( \cot \theta =\frac{2}{3} = \frac{2k}{3k} \)

Let us draw a \(\bigtriangleup ABC\) in which \(\angle B=90^{\circ}\) and \(\angle A = \theta \)

 

By pythagoras theorem, we have

\(AC^{2} = AB^{2} + BC^{2}\)
\(AC^{2} = (2k)^{2} + (3k)^{2}\)
\(AC^{2} = (4 k)^{2} + 9k^{2} = ??13k^{2} \)
\( AC = \sqrt{13k^{2}} = \sqrt{13}k \)
\( \sin \theta = \frac{3k}{\sqrt{13}k} = \frac{3}{\sqrt{13}} \)
\( \cos \theta = \frac{2k}{\sqrt{13}k} = \frac{2}{\sqrt{13}} \)

L.H.S.= \(= \frac{4\sin \theta -3\cos \theta }{2\sin \theta +6\cos \theta }\)

\(= \frac{4\times \frac{3}{\sqrt{13}} -3 \times \frac{2}{\sqrt{13}} }{2\times \frac{3}{\sqrt{13}} +6\times \frac{2}{\sqrt{13}} }\)
\(= \frac{\frac{12-6}{\sqrt{13}}}{\frac{6+12}{\sqrt{13}}}\)

\(= \frac{6}{18}=\frac{1}{3}\) = R.H.S.

Q7:

If \(3cot\Theta =4,\:show\:that\:\frac{\left ( 1-tan^{2}\Theta \right )}{\left ( 1+tan^{2}\Theta \right ) }=\left ( cos^{2} \Theta -sin^{2}\Theta \right )\).

Sol:

L.H.S. = \(\frac{\left ( 1-tan^{2}\Theta \right )}{\left ( 1+tan^{2}\Theta \right ) }\)

\(=\frac{\left ( 1-\frac{1}{cot^{2}\Theta } \right )}{\left ( 1+\frac{1}{cot^{2}\Theta } \right )}\)
\(=\frac{\left ( \frac{cot^{2}\Theta -1}{cot^{2}\Theta } \right )}{\left ( \frac{cot^{2}\Theta +1}{cot^{2}\Theta } \right )}\)
\(=\frac{cot^{2}\Theta -1}{cot^{2}\Theta +1}\)

\(=\frac{\left ( \frac{4}{3}\right )^{2}-1}{\left (\frac{4}{3} \right )^{2}+1}\) \(\left ( As,\:3cot\Theta =4\:or\:cot\Theta =\frac{4}{3} \right )\)

\(=\frac{\left ( \frac{16-9}{9} \right )}{\left ( \frac{16+9}{9} \right )}\)
\(=\frac{\left ( \frac{7}{9} \right )}{\left ( \frac{25}{9} \right )}\)
\(=\frac{7}{25}\)

R.H.S = \(\left ( cos^{2} \Theta -sin^{2}\Theta \right )\)

\(=\frac{\left ( cos^{2}\Theta -sin^{2}\Theta \right )}{1}\)
\(=\frac{\left ( \frac{cos^{2}\Theta -sin^{2}\Theta }{sin^{2}\Theta } \right )}{\left ( \frac{1}{sin^{2}\Theta } \right )}\)
\(=\frac{\left ( \frac{cos^{2}\Theta }{sin^{2}\Theta }-\frac{sin^{2}\Theta }{sin^{2}\Theta } \right )}{cosec^{2}\Theta }\)
\(=\frac{\left ( cot^{2}-1 \right )}{\left ( cot^{2}\Theta +1 \right )}\)
\(=\frac{\left [ \left ( \frac{4}{3} \right )^{2}-1 \right ]}{\left [ \left ( \frac{4}{3} \right )^{2}+1 \right ]}\)
\(=\frac{\left ( \frac{16-9}{9} \right )}{\left ( \frac{16+9}{9} \right )}\)
\(=\frac{\left ( \frac{7}{9} \right )}{\left ( \frac{25}{9} \right )}\)
\(=\frac{7}{25}\)

Since, L.H.S = R.H.S

Hence verified.

Q8:

\(If\:sec\Theta =\frac{17}{8},verify\:that\:\frac{\left ( 3-4sin^{2}\Theta \right )}{4cos^{2}\Theta -3}=\frac{\left ( 3-tan^{2} \Theta \right )}{\left ( 1-3tan^{2}\Theta \right ) }\)

Sol:

Given:

\(\sec \theta = \frac{17}{8} = \frac{17k}{8k} \)

Let us draw a \(\bigtriangleup ABC\) in which \(\angle B=90^{\circ} \) and \(\angle A = \theta \)

By pythagoras theorem, we have

\(AC^{2} = AB^{2} + BC^{2}\)
\(BC^{2} = AC^{2} – AB^{2}\)
\(BC^{2} = (17k)^{2} – (8k)^{2}\)
\(BC^{2} = 289k^{2} – 64k^{2} = 225k^{2}\)

Therefore, \(BC = \sqrt{225k^{2}} = 15k \)

\( \sin \theta = \frac{BC}{AC} = \frac{15k}{17k} \frac{15} {17} \)
\( \cos \theta = \frac{AB}{AC} = \frac{ 8k} {17k} = \frac{8}{17} \)
\( \tan \theta = \frac{15k} {8k} = \frac{15}{8} \)

L.H.S. = \(\frac{3-4\sin ^{2}\theta }{4\cos ^{2}\theta -3}= \frac{3-4\times \left ( \frac{15}{17} \right )^{2}}{4\times \left ( \frac{8}{17} \right )^{2} – 3}= \frac{3- \frac{4\times 225}{289}}{4\times \frac{64}{289}-3}\)

= \(\frac{3-4\sin ^{2}\theta }{4\cos ^{2}\theta -3}= \frac{3-4\times \left ( \frac{15}{17} \right )^{2}}{4\times \left ( \frac{8}{17} \right )^{2} – 3}= \frac{3- \frac{4\times 225}{289}}{4\times \frac{64}{289}-3}\)

R.H.S. = \(\frac{3 – \tan ^{2}\theta }{1 – 3\tan ^{2}\theta}= \frac{ 3 – \times \left ( \frac{15}{8} \right )^{2}}{1- 3\times \left ( \frac{15}{8} \right )^{2} }= \frac{3- \frac{225}{64}}{1 – 3\times \frac{225}{64}}= \frac{\frac{3\times 64- 225}{64}}{\frac{64 – 3\times 225}{64}}\)

\(= \frac{192-225}{64-675} = \frac{-33}{-611} = \frac{33}{611}\)

Q9:

In the adjoining figure, \(\angle B=90^{\circ},\angle BAC=\Theta ^{\circ}\), BC = CD = 4cm and AD = 10cm. Find \(\left ( i \right )sin\Theta \:and\:\left ( ii \right )cos\Theta\)

C:\Users\user\Desktop\1.PNG

Sol:

C:\Users\user\Desktop\1.PNG

In \(\Delta ABD\),

Using Pythagoras theorem, we get

\(AB=\sqrt{AD^{2}-BD^{2}}\)
\(=\sqrt{10^{2}-8^{2}}\)
\(=\sqrt{100-64}\)
\(=\sqrt{36}\)

= 6 cm

Again,

In \(\Delta ABC\),

Using Pythagoras theorem, we get

\(AC=\sqrt{AB^{2}+BC^{2}}\)
\(=\sqrt{6^{2}+4^{2}}\)
\(=\sqrt{36+16}\)
\(=\sqrt{52}\)
\(=2\sqrt{13}cm\)

Now,

(i) \(sin\Theta =\frac{BC}{AC}\)

\(=\frac{4}{2\sqrt{13}}\)
\(=\frac{2}{\sqrt{13}}\)
\(=\frac{2\sqrt{13}}{13}\)

(ii) \(cos\Theta =\frac{AB}{AC}\)

\(=\frac{6}{2\sqrt{13}}\)
\(=\frac{3}{\sqrt{13}}\)
\(=\frac{3\sqrt{13}}{13}\)

Q10:

In a \(\Delta ABC, \angle B=90^{\circ},AB=24cm\;and\;BC=7cm.\)

Find (i) sin A (ii) cos A (iii) sin C (iv) cos C

Sol:

Given:

\(\bigtriangleup ABC \) in which \(\angle B = 90^{\circ} \), AB= 24 cm, BC = 7 cm

 

By pythagoras theorem, we have

\(AC^{2} = 24^{2} + 7^{2} \; cm^{2} = ( 576 + 49 ) \; cm^{2} \)
\( = 625 \; cm^{2} \)
\( = 25 \; cm^{2} \)

Therefore, \(AC = 25 \) cm

(i) For T-ratio of \(\angle A\), we have

Base = AB = 24 cm

Perpendicular = BC = 7 cm and

Hypotenuse = AC = 25 cm

\( \cos A = \frac{AB}{AC}= \frac{24}{25}\) and

\( \sin A= \frac{BC}{AC}= \frac{7}{25} \)

(ii) For T-ratio of \(\angle C\), we have

Base = BC = 7 cm

Perpendicular = AB = 24 cm and

Hypotenuse = AC = 25 cm

\( \cos C = \frac{BC}{AC}= \frac{7}{25} \) and

\( \sin C = \frac{AB} {AC}= \frac{24} {25}\)

Q11:

In a \(\Delta ABC,\angle C=90^{\circ},\angle ABC=\Theta ,BC=21\) units and AB = 29 units.

Show that \(\left ( cos^{2}\Theta -sin^{2} theta\right )=\frac{41}{841}\)

Sol:

Given:

\(\bigtriangleup ABC \) in which \(\angle B = 90^{\circ} \; \angle A = \theta \), BC = 21 units, AB = 29 units

(i) \((\cos ^{2}\theta +\sin ^{2}\theta )= \left ( \frac{21}{29} \right )^{2}+ \left ( \frac{20}{29} \right )^{2}\)

\(= \left ( \frac{441}{841} \right )+ \left ( \frac{400}{841} \right )\)
\(= \left ( \frac{841}{841} \right ) = 1 \)
\((\cos ^{2}\theta +\sin ^{2}\theta )= 1 \)

(ii) \((\cos ^{2}\theta – \sin ^{2}\theta )= \left ( \frac{21}{29} \right )^{2} – ??\left ( \frac{20}{29} \right )^{2}\)

\(= \left ( \frac{441}{841} \right ) – \left ( \frac{400}{841} \right )\)
\(= \left ( \frac{41}{841} \right ) = 1 \)

Hence, ??\((\cos ^{2}\theta – \sin ^{2}\theta ) = 1 \)

Q12:

In a \(\Delta ABC,\angle B=90^{\circ},AB=12cm\:and\:BC=5cm.\)

Find (i) cos A (ii) cosec A (iii) cos C (iv) cosec C

Sol:

Given:

\(\bigtriangleup ABC \) in which \(\angle B = 90^{\circ}\),

AB= 12 cm, BC= 5cm

By pythagoras theorem, we have

\(AC^{2} = AB^{2} + BC^{2} = (12)^{2} + (5)^{2} \)
\( = 144k^{2} + 25^{2} \)
\( = 169^{2} \)

Therefore, \(AC = 13 \)

For T-ratio of \(\angle A\), we have

Base = AB = 12 cm

Perpendicular = BC = 5 cm and

Hypotenuse = AC = 13 cm

Therefore, \(\cos A= \frac{AB}{AC}= \frac{12}{13}\) and

Therefore, \(cosec A= \frac{AC}{BC}= \frac{13}{5}\)

For T-ratio of \(\angle C\), we have

Base = BC = 5 cm

Perpendicular = AB = 12 cm and

Hypotenuse = AC = 13 cm

\( \cos C = \frac{BC}{AC}= \frac{5}{13} \) and

\( cosec C = \frac{AC} {BC}= \frac{13} {5}\)

Q13:

If \(sin\alpha =\frac{1}{2},prove\:that\:\left ( 3cos\alpha -4cos^{3} \alpha \right )=0\)

Sol:

L.H.S = \(( 3cos\alpha -4cos^{3} \alpha )\)

\(=cos\alpha \left ( 3-4cos^{3}\alpha \right )\)
\(=\sqrt{1-sin^{2}\alpha }\left [ 3-4\left ( 1-sin^{2}\alpha \right ) \right ]\)
\(=\sqrt{1-\left ( \frac{1}{2} \right )^{2}}\left [ 3-4\left ( 1-\left ( \frac{1}{2} \right )^{2} \right ) \right ]\)
\(=\sqrt{\frac{1}{1}-\frac{1}{4}}\left [ 3-4\left ( \frac{1}{1}-\frac{1}{4} \right ) \right ]\)
\(=\sqrt{\frac{3}{4}}\left [ 3-4\left ( \frac{3}{4} \right ) \right ]\)
\(=\sqrt{\frac{3}{4}}\left [ 3-3 \right ]\)
\(=\sqrt{\frac{3}{4}}\left [ 0 \right ]\)

= 0

= R. H. S

Q14:

In a \(\Delta ABC,\angle B=90^{\circ}\:and\:tanA=\frac{1}{\sqrt{3}}.\:\:Prove\:that\:\left ( i \right )sinA.cosC+cosA.sinC=1\:\:\left ( ii \right )cosA.cosC-sinA.sinC=0\)

Sol:

C:\Users\user\Desktop\2.PNG

In \(\Delta ABC,\angle B=90^{\circ}\),

As, tanA = \(\frac{1}{\sqrt{3}}\)

\(=\frac{BC}{AB}=\frac{1}{\sqrt{3}}\)

Let BC = x and AB = \(x\sqrt{3}\)

Using Pythagoras theorem, we get

\(AC=\sqrt{AB^{2}+BC^{2}}\)
\(=\sqrt{\left ( X\sqrt{3} \right )^{2}+x^{2}}\)
\(=\sqrt{3x^{2}+x^{2}}\)
\(=\sqrt{4x^{2}}\)

= 2x

Now,

(i) L.H.S = sinA.cosC + cosA.sinC

\(=\frac{BC}{AC}.\frac{BC}{AC}+\frac{AB}{AC}.\frac{AB}{AC}\)
\(=\left ( \frac{BC}{AC} \right )^{2}+\left ( \frac{AB}{AC} \right )^{2}\)
\(=\left ( \frac{x}{2x} \right )^{2}+\left ( \frac{x\sqrt{3}}{2x} \right )^{2}\)
\(=\frac{1}{4}+\frac{3}{4}\)
\(=\frac{4}{4}\)

= 1

= R.H.S

(ii) L.H.S = cosA.cosC ??? sinA.sinC

\(=\frac{AB}{AC}.\frac{BC}{AC}-\frac{BC}{AC}.\frac{AB}{AC}\)
\(=\frac{x\sqrt{3}}{2x}.\frac{x}{2x}-\frac{x}{2x}.\frac{x\sqrt{3}}{2x}\)
\(=\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}\)

= 0

= R.H.S

Q15:

If \(\angle A\:and\:\angle B\) are acute angles such that sinA = sinB then prove that \(\angle A=\angle B\).

Sol:

C:\Users\user\Desktop\3.PNG

In \(\Delta ABC,\angle B=90^{\circ}\),

sinA = \(\frac{BC}{AB}\) and

sinB = \(\frac{AC}{AB}\)

As, sinA = sinB

\(\Rightarrow \frac{BC}{AB}=\frac{AC}{AB}\)

So, \(\angle A=\angle B\) (Angles opposite to each sides are equal)

Q16:

If \(\angle A\:and\:\angle B\) are acute angles such that tanA = tanB then prove that \(\angle A=\angle B\).

Sol:

C:\Users\user\Desktop\3.PNG

In \(\Delta ABC,\angle C=90^{\circ}\),

tanA = \(\frac{BC}{AC}\) and

tanB = \(\frac{AC}{BC}\)

As, tanA = tanB

\(\Rightarrow \frac{BC}{AC}=\frac{AC}{BC}\)
\(\Rightarrow BC^{2}=AC^{2}\)

=> BC = AC

So, \(\angle A=\angle B\) (Angles opposite to each sides are equal)

Q17:

In a right \(\Delta ABC\), right-angled at B, if tanA = 1 then verify that 2sinA.cosA = 1.

Sol:

We have,

tanA = 1

=> \(\Delta ABC\) = 1

=>sinA = cosA

=>sinA – cosA = 0

Squaring both sides, we get

\(\left ( sinA-cosA \right )^{2}=0\)
\(\Rightarrow sin^{2}A+cosA^{2}-2sinA.cosA=0\)

=> 1- 2sinA.cosA =0

Therefore, \(2sinA.cosA=1\)

Q18:

In the figure of \(\Delta PQR,\angle P=\Theta ^{\circ}\:and\:\angle R=\phi ^{\circ}.\)

C:\Users\user\Desktop\4.PNG

Find:

(i) \(\left ( \sqrt{x+1} \right )cot\phi\)

(ii) \(\left ( \sqrt{x^{3}+x^{2}} \right )tan\Theta\)

(iii) \(cos\Theta\)

Sol:

C:\Users\user\Desktop\4.PNG

In \(\Delta PQR,\angle Q=90^{\circ}\),

Using Pythagoras theorem, we get

\(PQ=\sqrt{PR^{2}-QR^{2}}\)
\(=\sqrt{\left ( x+2 \right )^{2}-x^{2}}\)
\(=\sqrt{x^{2}+4x+4-x^{2}}\)
\(=\sqrt{4\left ( x+1 \right )}\)
\(=2\sqrt{x+1}\)

Now,

(i)\(\left ( \sqrt{x+1} \right )cot\phi\)

\(=\left ( \sqrt{x+1} \right )\times \frac{QR}{PQ}\)
\(=\left ( \sqrt{x+1} \right )\times \frac{x}{2\sqrt{x+1}}\)
\(=\frac{x}{2}\)

(ii)\(\left ( \sqrt{x^{3}+x^{2}} \right )tan\Theta\)

\(=\left ( \sqrt{x^{2}\left ( x+1 \right )} \right )\times \frac{QR}{PQ}\)
\(=x\sqrt{\left ( x+1 \right )}\times \frac{x}{2\sqrt{x+1}}\)
\(=\frac{x^{2}}{2}\)

(iii)\(cos\Theta\)

\(=\frac{PQ}{PR}\)
\(=\frac{2\sqrt{x+1}}{\left ( x+2 \right )}\)

Q19:

If x = cosecA + cosA and y = cosecA ??? cosA then prove that

\(\left ( \frac{2}{x+y} \right )^{2}+\left ( \frac{x-y}{2} \right )^{2}-1=0\)

Sol:

L.H.S = \(\left ( \frac{2}{x+y} \right )^{2}+\left ( \frac{x-y}{2} \right )^{2}-1\)

\(=\left [ \frac{2}{\left ( cosecA+cosA \right )+\left ( cosecA-cosA \right )} \right ]^{2}+\left [ \frac{\left ( cosecA+cosA \right )-\left ( cosecA-cosA \right )}{2} \right ]^{2}-1\)
\(=\left [ \frac{2}{cosecA+cosA+cosecA-cosA} \right ]^{2}+\left [ \frac{cosecA+cosA-cosecA+cosA}{2} \right ]^{2}-1\)
\(=\left [ \frac{2}{2cosecA} \right ]^{2}+\left [ \frac{2cosA}{2} \right ]^{2}-1\)
\(=\left [ \frac{1}{cosecA} \right ]^{2}+\left [ cosA \right ]^{2}-1\)
\(=\left [ sinA \right ]^{2}+\left [ cosA \right ]^{2}-1\)
\(=sin^{2}A+cos^{2}A-1\)

= 1-1 = 0

= 0

= R.H.S

Q20:

If x = cot A + cos A and y = cot A ??? cos A, prove that

\(\left ( \frac{x-y}{x+y} \right )^{2}+\left ( \frac{x-y}{2} \right )^{2}=1\)

Sol:

L.H.S = \(\left ( \frac{x-y}{x+y} \right )^{2}+\left ( \frac{x-y}{2} \right )^{2}\)

\(=\left [ \frac{\left ( cotA+cosA \right )-\left ( cotA-cosA \right )}{\left ( cotA+cosA \right )+\left ( cotA-cosA \right )} \right ]^{2}+\left [ \frac{\left ( cotA+cosA \right )-\left ( cotA-cosA \right )}{2} \right ]^{2}\)
\(=\left [ \frac{cotA+cosA-cotA+cosA}{cotA+cosA+cotA-cosA} \right ]+\left [ \frac{cotA+cosA-cotA+cosA}{2} \right ]^{2}\)
\(=\left [ \frac{2cosA}{2cotA} \right ]^{2}+\left [ \frac{2cosA}{2} \right ]^{2}\)
\(=\left [ \frac{cosA}{\left ( \frac{cosA}{sinA} \right )} \right ]^{2}+\left [ cosA \right ]^{2}\)
\(=\left [ \frac{sinA.cosA}{cosA} \right ]^{2}+\left [ cosA \right ]^{2}\)
\(=\left [ sinA \right ]^{2}+\left [ cosA \right ]^{2}\)
\(=sin^{2}A+cos^{2}A\)

= 1

= R.H.S


Practise This Question

A ball is connected to a rope and swung around in uniform circular motion. The tension in the rope is measured at 10 N and the radius of the circle is 1 m. How much work is done in one revolution around the circle?