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This is one of our many quizzes on Trigonometric Identities. Click on this to access more such competitive quizzes on Trigonometric Functions.

Trigonometric Identities Quiz 2

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 If  $p\,=\,{{\sin }^{2}}\,\dfrac{3m}{10}+\,{{\cos }^{2}}\dfrac{3m}{10}$ and  $q\,=\,Co\,se{{c}^{2}}\dfrac{7m}{9}-\,{{\cot }^{2}}\dfrac{7m}{9}$, then write the value of ${{\left( {{p}^{3}}\,+\,{{q}^{3}} \right)}^{3}}$.

 Use the identities ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ and $\cos e{{c}^{2}}\theta -{{\cot }^{2}}=1$.

  $\begin{align}  & \text{Consider the given values of }p\text{ and }q \\  & p={{\sin }^{2}}\left( \dfrac{3m}{10} \right)+{{\cos }^{2}}\left( \dfrac{3m}{10} \right)=1\text{    }..........\left( \text{i} \right) \\  & q=\cos e{{c}^{2}}\left( \dfrac{7m}{9} \right)-{{\cot }^{2}}\left( \dfrac{7m}{9} \right)=1\text{  }.........\left( \text{ii} \right) \\  & \text{ }\!\![\!\!\text{ using }\left( \text{i} \right)\text{ and }\left( \text{ii} \right)\text{ }\!\!]\!\!\text{ } \\  & \Rightarrow {{p}^{3}}+{{q}^{3}}={{\left( 1 \right)}^{3}}+{{\left( 1 \right)}^{3}}=2 \\  & \Rightarrow {{\left( {{p}^{3}}+{{q}^{3}} \right)}^{3}}={{\left( 2 \right)}^{3}}=8 \\ \end{align}$

 Write the value of ${{\left( 12 \right)}^{z}}$ where $z=\sqrt{\dfrac{1-\cos \dfrac{17m}{5}}{1+\cos \dfrac{17m}{5}}}$   at $m=0$.

 Use trigonometric identities $1-cos^2\theta=2sin^2\theta$ and $1+cos^2\theta=2cos^2\theta$ to solve the expression.

  $\begin{align}  & \text{Given,} \\  & z=\sqrt{\dfrac{1-\cos \left( \dfrac{17m}{5} \right)}{1+\cos \left( \dfrac{17m}{5} \right)}} \\  & \Rightarrow \sqrt{\dfrac{2{{\sin }^{2}}\dfrac{\left( \dfrac{17m}{5} \right)}{2}}{2{{\cos }^{2}}\dfrac{\left( \dfrac{17m}{5} \right)}{2}}}=\sqrt{\dfrac{2{{\sin }^{2}}\left( \dfrac{17m}{10} \right)}{2{{\cos }^{2}}\left( \dfrac{17m}{10} \right)}} \\  & \Rightarrow z=\tan \dfrac{17m}{10}\text{   }.........\left( \text{i} \right) \\  & \text{Substitute }m=0\text{ }~\text{ }\!\!~\!\!\text{ in equation }\left( \text{i} \right) \\  & z=\tan \left( \dfrac{17\times 0}{10} \right)=0 \\  & \Rightarrow {{12}^{z}}={{12}^{\circ}}=1 \\ \end{align}$

 Evaluate ${{\left[ \cot \,\left( \dfrac{29\,w}{4} \right)+\,\tan \,\left( \dfrac{29\,w}{4} \right) \right]}^{2}}$ and write its simplest form.

 Use $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$ and $tan\theta =\dfrac{\sin \theta }{\cos \theta }$. Then take L.C.M. and solve further.

$\begin{align} &nbsp;&nbsp;& \text{Write the given expression,} \\ &nbsp;& {{\left[ \cot \left( \dfrac{29w}{4} \right)+\tan \left( \dfrac{29w}{4} \right) \right]}^{2}} \\ &nbsp;& ={{\left[ \dfrac{\cos \left( \dfrac{29w}{4} \right)}{\sin \left( \dfrac{29w}{4} \right)}+\dfrac{\sin \left( \dfrac{29w}{4} \right)}{\cos \left( \dfrac{29w}{4} \right)} \right]}^{2}} \\ &nbsp;& ={{\left[ \dfrac{{{\cos }^{2}}\left( \dfrac{29w}{4} \right)+{{\sin }^{2}}\left( \dfrac{29w}{2} \right)}{\sin \left( \dfrac{29w}{4} \right)\cos \left( \dfrac{29w}{4} \right)} \right]}^{2}} \\ &nbsp;& ={{\left[ \dfrac{1}{\sin \left( \dfrac{29w}{4} \right)\cos \left( \dfrac{29w}{4} \right)} \right]}^{2}}={{\left[ \sec \left( \dfrac{29w}{4} \right)\cos ec\left( \dfrac{29w}{4} \right) \right]}^{2}} \\ &nbsp;& ={{\sec }^{2}}\left( \dfrac{29w}{4} \right)\cos e{{c}^{2}}\left( \dfrac{29w}{4} \right) \\ \end{align}$

 Write the value of ${{e}^{{{\left( \sin \,\dfrac{h}{9}+\cos \dfrac{h}{9} \right)}^{2}}}}$.

 Use the identities ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ and ${{\sin }{2}}\theta =2\sin \theta \cos \theta $ to simplify the given expression. 

$\begin{align} &nbsp;&nbsp;& \text{Consider the given expression,} \\ &nbsp;& {{e}^{{{\left( \sin \dfrac{h}{9}+\cos \dfrac{h}{9} \right)}^{2}}}} \\ &nbsp;& ={{e}^{{{\sin }^{2}}\left( \dfrac{h}{9} \right)+2\sin \left( \dfrac{h}{9} \right)\cos \left( \dfrac{h}{9} \right)+{{\cos }^{2}}\left( \dfrac{h}{9} \right)}} \\ &nbsp;& ={{e}^{1+2\sin \left( \dfrac{h}{9} \right)\cos \left( \dfrac{h}{9} \right)}} \\ &nbsp;& \left[ \because {{\sin }^{2}}\dfrac{h}{9}+{{\cos }^{2}}\dfrac{h}{9}=1 \right] \\ &nbsp;& \Rightarrow {{e}^{1+\sin \left( \dfrac{2h}{9} \right)}} \\ &nbsp;& \left[ \because 2\sin \theta \cos \theta =\sin 2\theta \right] \\ \end{align}$.

 Simplify: $\dfrac{Co\,{{\sec }^{12}}\left( \dfrac{5\ell }{3} \right)\,{{\cot }^{10}}\,\left( \dfrac{5\ell }{3} \right)\,-\,Co\,{{\sec }^{10}}\,\left( \dfrac{5\ell }{3} \right)\,{{\cot }^{12}}\,\left( \dfrac{5\ell }{3} \right)}{Co\,{{\sec }^{5}}\left( \dfrac{5\ell }{3} \right)\,{{\cot }^{5}}\left( \dfrac{5\ell }{3} \right)}$.

 Take common in the numerator and use the trigonometric identity $\cos e{{c}^{2}}\theta -{{\cot }^{2}}\theta =1$.

$\begin{align} &nbsp;&nbsp;& \text{Consider the given expression,} \\ &nbsp;& \dfrac{Co{{\sec }^{12}}\left( \dfrac{5\ell }{3} \right){{\cot }^{10}}\left( \dfrac{5\ell }{3} \right)-\cos e{{c}^{10}}\left( \dfrac{5\ell }{3} \right){{\cot }^{12}}\left( \dfrac{5\ell }{3} \right)}{\cos e{{c}^{5}}\left( \dfrac{5\ell }{3} \right){{\cot }^{5}}\left( \dfrac{5\ell }{3} \right)} \\ &nbsp;& \text{Take }\cos e{{c}^{10}}\left( \dfrac{5\ell }{3} \right){{\cot }^{10}}\left( \dfrac{5\ell }{3} \right)\text{ common is the numerator,} \\ &nbsp;& \Rightarrow \dfrac{\cos e{{c}^{10}}\left( \dfrac{5\ell }{3} \right){{\cot }^{10}}\left( \dfrac{5l}{3} \right)\left[ \cos e{{c}^{2}}\left( \dfrac{5\ell }{3} \right)-{{\cot }^{2}}\left( \dfrac{5\ell }{3} \right) \right]}{\cos e{{c}^{5}}\left( \dfrac{5\ell }{3} \right){{\cot }^{5}}\left( \dfrac{5\ell }{3} \right)} \\ &nbsp;& =\left[ \cos e{{c}^{2}}\left( \dfrac{5\ell }{3} \right)-{{\cot }^{2}}\left( \dfrac{5\ell }{3} \right) \right]\cos e{{c}^{5}}\left( \dfrac{5\ell }{3} \right){{\cot }^{5}}\left( \dfrac{5\ell }{3} \right) \\ &nbsp;& =1.\cos e{{c}^{5}}\left( \dfrac{5\ell }{3} \right){{\cot }^{5}}\left( \dfrac{5\ell }{3} \right) \\ &nbsp;& =\cos e{{c}^{5}}\left( \dfrac{5\ell }{3} \right).{{\cot }^{5}}\left( \dfrac{5\ell }{3} \right) \\ \end{align}$