RD Sharma Solutions Class 10 Trigonometric Identities Exercise 6.2

RD Sharma Solutions Class 10 Chapter 6 Exercise 6.2

RD Sharma Class 10 Solutions Chapter 6 Ex 6.2 PDF Download

Exercise 6.2

 

 Q1) If \(cos\theta =\frac{4}{5}\), find all other trigonometric ratios of angle \(\Theta\).

 

Solution:

We have:

\(sin\Theta=\sqrt{1-cos^{2}\Theta}=\sqrt{1-(\frac{4}{5})^{2}}\)

= \(\sqrt{1-\frac{16}{25}}\)

= \(\sqrt{\frac{25-16}{25}}\)

= \(\sqrt{\frac{9}{25}}=\frac{3}{5}\)

Therefore, \(sin\Theta=\frac{3}{5}\)

\(tan\Theta=\frac{sin\Theta}{cos\Theta}=\frac{\frac{3}{5}}{\frac{4}{5}}=\frac{3}{4}sec\Theta=\frac{1}{cos\Theta}=\frac{1}{\frac{4}{5}}=\frac{5}{4}\)

i.e. \(cosec\Theta=\frac{1}{sec\Theta}=\frac{1}{\frac{3}{5}}=\frac{5}{3}cot\Theta=\frac{1}{tan\Theta}=\frac{1}{\frac{3}{4}}=\frac{4}{3}\)

 

Q2) If \(sin\Theta =\frac{1}{\sqrt{2}}\), find all other trigonometric ratios of angle \(\Theta\).

 

Solution:

 We have,

\(cos\Theta=\sqrt{1-sin^{2}\Theta}=\sqrt{1-(\frac{1}{\sqrt{2}})^{2}}\)

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

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

= \(cos\Theta=\frac{1}{\sqrt{2}}\)

= \(tan\Theta=\frac{sin\Theta}{cos\Theta}=\frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}=1\)

= \(cosec\Theta=\frac{1}{sin\Theta}=\frac{1}{\frac{1}{\sqrt{2}}}=\sqrt{2}\)

= \(sec\Theta=\frac{1}{cos\Theta}=\frac{1}{\frac{1}{\sqrt{2}}}=\sqrt{2}\)

= \(cot\Theta=\frac{1}{tan\Theta}=\frac{1}{1}=1\)

 

Q3) If \(tan\Theta=\frac{1}{\sqrt{2}}\), find the value of \(\frac{cosec^{2}\Theta-sec^{2}\Theta}{cosec^{2}\Theta+cot^{2}\Theta}\).

 

Solution:

We know that \(sec\Theta =\sqrt{1+tan^{2}\Theta}\)

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

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

= \(cot\Theta=\frac{1}{tan\Theta}=\frac{1}{\frac{1}{\sqrt{2}}}=\sqrt{2}\)

= \(cosec\Theta=\sqrt{1+cot^{2}\Theta}=\sqrt{1+2}=\sqrt{3}\)

Substituting it in equation (1) we get

= \(\frac{(\sqrt{3})^{2}-(\sqrt{\frac{3}{2}})^{2}}{(\sqrt{3})^{2}+(\sqrt{2})^{2}}=\frac{3-\frac{3}{2}}{3+2}=\frac{\frac{3}{2}}{5}=\frac{3}{10}\)

 

Q4) If \(tan\Theta=\frac{3}{4}\), find the value of \(\frac{1-cos\Theta}{1+cos\Theta}\)

 

Solution:

We know that

\(sec\Theta =\sqrt{1+tan^{2}\Theta}\)

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

= \(\sqrt{1+\frac{9}{16}}\)

=\(\sqrt{\frac{16+9}{16}}\)

= \(\sqrt{\frac{25}{16}}\)

= \(sec\Theta=\frac{5}{4}\)

= \(sec\Theta=\frac{1}{cos\Theta}=\frac{1}{\frac{5}{4}}=\frac{4}{5}=cos\Theta\)

Therefore, \(We\;get\;\frac{1-\frac{4}{5}}{1+\frac{4}{5}}=\frac{\frac{1}{5}}{\frac{9}{5}}=\frac{1}{9}\)

 

Q5) If \(tan\Theta=\frac{12}{5}\), find the value of \(\frac{1+sin\Theta}{1-sin\Theta}\).

 

Solution:

\(cot\Theta =\frac{1}{tan\Theta}=\frac{1}{\frac{12}{5}}=\frac{5}{12}\)

= \(cosec\Theta=\sqrt{1+cot^{2}\Theta}=\sqrt{1+[\frac{5}{12}]^{2}}=\sqrt{\frac{144+25}{144}}=\sqrt{\frac{169}{144}}=\frac{13}{12}\)

= \(sin\Theta=\frac{1}{cosec\Theta}=\frac{1}{\frac{13}{12}}=\frac{12}{13}\)

i.e. \(We\;get\;\frac{1+\frac{12}{13}}{1-\frac{12}{13}}=\frac{\frac{13+12}{18}}{\frac{13-12}{18}}=\frac{25}{1}=25\).

 

Q6) If \(cot\Theta=\frac{1}{\sqrt{3}}\), find the value of \(\frac{1-cos^{2}\Theta}{2-sin^{2}\Theta}\).

 

Solution:

\(cosec\Theta =\sqrt{1+cot^{2}\Theta}=\sqrt{1+\frac{1}{3}}=\sqrt{\frac{4}{3}}\)

= \(cosec\Theta=\frac{2}{\sqrt{3}}\)

= \(sin\Theta=\frac{1}{cosec\Theta}=\frac{1}{\frac{2}{\sqrt{3}}}=\frac{\sqrt{3}}{2}\)

= \(and\;\frac{1}{cot\Theta}=\frac{sin\Theta}{cos\Theta}=cos\Theta=sin\Theta\times cot\Theta=\frac{\sqrt{3}}{2}\times \frac{1}{\sqrt{3}}=\frac{1}{2}\)

Therefore, on substituting we get

= \(\frac{1-(\frac{1}{2})^{2}}{2-(\frac{\sqrt{3}}{2})^{2}}=\frac{1-\frac{1}{4}}{2-\frac{3}{4}}=\frac{\frac{3}{4}}{\frac{5}{4}}=\frac{3}{5}\).

 

Q7) If \(cosecA=\sqrt{2}\), find the value of \(\frac{2sin^{2}A+3cot^{2}A}{4(tan^{2}A-cos^{2}A)}\).

 

Solution:

We know that \(cotA=\sqrt{cosec^{2}A-1}\)

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

= \(tanA=\frac{1}{cotA}=\frac{1}{1}=1\)

= \(sinA=\frac{1}{cosecA}=\frac{1}{\sqrt{2}}\)

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

\(cosA=\sqrt{1-sin^{2}A}=\sqrt{1-(\frac{1}{\sqrt{2}})^{2}}=\sqrt{\frac{1}{\sqrt{2}}}=\frac{1}{\sqrt{2}}\)

On substituting we get:

= \(\frac{2[\frac{1}{\sqrt{2}}]^{2}+3[1]^{2}}{4[[1]-[\frac{1}{\sqrt{2}}]^{2}]}=\frac{2\times \frac{1}{2}+3}{4[1-\frac{1}{2}]}\)

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

 

Q8) If  \(cot\Theta=\sqrt{3}\), find the value of \(\frac{cosec^{2}\Theta+cot^{2}\Theta}{cosec^{2}\Theta-sec^{2}\Theta}\).

 

Solution:

\(cosec\Theta=\sqrt{1+cot^{2}\Theta}=\sqrt{1+(\sqrt{3})^{2}}=\sqrt{1+3}=2\)

\(sin\Theta=\frac{1}{cosec\Theta}=\frac{1}{2}cot\Theta=\frac{cos\Theta}{sin\Theta}\;\;\;\;\;\;\cos\Theta=cot\Theta.sin\Theta\)

\(\Rightarrow cos\Theta=\frac{\sqrt{3}}{2}\)

= \(sec\Theta=\frac{1}{cos\Theta}=\frac{2}{\sqrt{3}}\)

On substituting we get:

\(\frac{(2)^{2}+(\sqrt{3})^{2}}{(2)^{2}-(\frac{2}{\sqrt{3}})^{2}}=\frac{4+3}{\frac{12-4}{3}}=\frac{7}{\frac{8}{3}}\)

= \(\frac{21}{8}\)

 

Q9) If \(3cos\Theta=1\), find the value of \(\frac{6sin^{2}\Theta+tan^{2}\Theta}{4cos\Theta}\).

 

Solution:

\(cos\Theta=\frac{1}{3},\;\;\;\;\;\;\;sin\Theta=\sqrt{1-cos^{2}\Theta}\)

= \(\sqrt{1-\frac{1}{9}}=\sqrt{\frac{8}{9}}=\frac{2\sqrt{2}}{3}\)

\(tan\Theta=\frac{sin\Theta}{cos\Theta}=\frac{2\sqrt{2}}{3.\frac{1}{3}}=2\sqrt{2}\)

On substituting we get

\(\frac{6[\frac{2\sqrt{2}}{3}]^{2}+(2\sqrt{2})^{2}}{4.\frac{1}{3}}=\frac{\frac{16}{3}+8}{\frac{4}{3}}=\frac{\frac{16+24}{3}}{\frac{4}{3}}\)

= \(\frac{40}{4}=10\)

 

Q10) If \(\sqrt{3}tan\Theta=sin\Theta\), find the value of \(sin^{2}\Theta-cos^{2}\Theta\).

 

Solution:

\(\sqrt{3}\frac{sin\Theta}{cos\Theta}=sin\Theta\)

= \(cos\Theta=\frac{\sqrt{3}}{3}\Rightarrow \frac{1}{\sqrt{3}}\)

= \(sin\Theta=\sqrt{1-cos^{2}\Theta}=\sqrt{1-(\frac{1}{\sqrt{3}})^{2}}\)

= \(sin^{2}\Theta-cos^{2}\Theta=(\sqrt{\frac{2}{3}})^{2}-(\frac{1}{\sqrt{3}})^{2}\)

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

 

Q11) If \(cosec\Theta=\frac{13}{12}\), find the value of \(\frac{2sin\Theta-3cos\Theta}{4sin\Theta-9cos\Theta}\).

 

Solution:

\(sin\Theta=\frac{1}{cosec\Theta}=\frac{1}{\frac{13}{12}}=\frac{12}{13}\)

= \(cos\Theta=\sqrt{1-sin^{2}\Theta}=\sqrt{1-[\frac{12}{13}]^{2}}=\sqrt{1-\frac{144}{169}}\)

= \(\sqrt{\frac{25}{169}}=\frac{5}{13}\)

\(\Rightarrow \frac{2.\frac{12}{13}-3.\frac{5}{13}}{4.\frac{12}{13}-9.\frac{5}{13}}=\frac{\frac{24-15}{13}}{\frac{48-15}{13}}=\frac{9}{3}=3\)

 

Q12) If \(sin\Theta+cos\Theta=\sqrt{2}cos(90^{\circ}-\Theta)\), find \(cot\Theta\).

 

Solution:

= \(sin\Theta+cos\Theta=\sqrt{2}sin\Theta\;\;\;\;\;\;[cos(90-\Theta)=sin\Theta]\)

\(\Rightarrow cos\Theta=\sqrt{2}sin\Theta-sin\Theta\\ \Rightarrow cos\Theta=sin\Theta(\sqrt{2}-1)\)

Divide both sides with \(sin\Theta\) we get

= \(\frac{cos\Theta}{sin\Theta}=\frac{sin\Theta}{sin\Theta}(\sqrt{2}-1)\)

= \(cot\Theta=\sqrt{2}-1\).

 

Q-13. If \(2sin^{2}\Theta – cos^{2}\Theta = 2\), then find the value of \(\Theta\).

 

Solution.

\(2sin^{2}\Theta – cos^{2}\Theta = 2\)

\(\Rightarrow  2sin^{2}\Theta -\left ( 1 – sin^{2}\Theta \right ) = 2\)

\(\Rightarrow 2sin^{2}\Theta – 1 + sin^{2}\Theta = 2\)

\(\Rightarrow 3sin^{2}\Theta = 3\)

\(\Rightarrow sin^{2}\Theta = 1\)

\(\Rightarrow sin \Theta = 1\)

\(\Rightarrow sin\Theta = sin 90^{\circ}\)

\(\Rightarrow \Theta = 90^{\circ}\)

 

Q-14. If \(\sqrt{3} tan\Theta – 1 = 0\), find the value of \(sin^{2}\Theta – cos^{2}\Theta\).

 

Solution.

\(\sqrt{3} tan\Theta – 1 = 0\)

\(\Rightarrow  \sqrt{3} tan\Theta = 1\)

\(\Rightarrow  \sqrt{3} tan\Theta = \frac{1}{\sqrt{3}}\)

\(\sqrt{3} tan\Theta = tan 30^{\circ}\)

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

Now,

\(sin^{2}\Theta – cos^{2}\Theta\)

= \(sin^{2}\left ( 30^{\circ} \right ) – cos^{2}\left ( 30^{\circ} \right )\)

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

= \(\frac{1}{4} – \frac{3}{4}\) = \(\frac{-2}{4}\) = \(\frac{-1}{2}\)