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This is one of our many quizzes on Simplifying trigonometric expressions. Click on this to access more such competitive quizzes on [Parent Topic Name].

Simplifying trigonometric expressions Quiz 2

mathematics

 Write the simplified form of the trigonometric expression $\dfrac{{{\tan }^{2}}\left( \dfrac{6m}{5} \right)-{{\sin }^{2}}\left( \dfrac{6m}{5} \right)}{{{\tan }^{2}}\left( \dfrac{6m}{5} \right)+{{\sin }^{2}}\left( \dfrac{6m}{5} \right)}$.

 Use trigonometric formulas $\tan x = \dfrac{\sin x}{\cos x}$ to reduce the expression into simplest form.

  $\begin{align}  & \text{Consider the given expression,} \\  & \dfrac{{{\tan }^{2}}\left( \dfrac{6m}{5} \right)-{{\sin }^{2}}\left( \dfrac{6m}{5} \right)}{{{\tan }^{2}}\left( \dfrac{6m}{5} \right)+{{\sin }^{2}}\left( \dfrac{6m}{5} \right)} \\  & =\dfrac{\dfrac{{{\sin }^{2}}\left( \dfrac{6m}{5} \right)}{{{\cos }^{2}}\left( \dfrac{6m}{5} \right)}-{{\sin }^{2}}\left( \dfrac{6m}{5} \right)}{\dfrac{{{\sin }^{2}}\left( \dfrac{6m}{5} \right)}{{{\cos }^{2}}\left( \dfrac{6m}{5} \right)}+{{\sin }^{2}}\left( \dfrac{6m}{5} \right)} \\  & =\dfrac{{{\sin }^{2}}\left( \dfrac{6m}{5} \right)\left[ \dfrac{1-{{\cos }^{2}}\left( \dfrac{6m}{5} \right)}{{{\cos }^{2}}\left( \dfrac{6m}{5} \right)} \right]}{{{\sin }^{2}}\left( \dfrac{6m}{5} \right)\left[ \dfrac{1+{{\cos }^{2}}\left( \dfrac{6m}{5} \right)}{{{\cos }^{2}}\left( \dfrac{6m}{5} \right)} \right]} \\  & =\dfrac{1-{{\cos }^{2}}\left( \dfrac{6m}{5} \right)}{1+{{\cos }^{2}}\left( \dfrac{6m}{5} \right)} \\  & =\dfrac{{{\sin }^{2}}\left( \dfrac{6m}{5} \right)}{1+{{\cos }^{2}}\left( \dfrac{6m}{5} \right)} \\ \end{align}$

 Evaluate the expression $\log\left(\dfrac{2^{\sec \:\:^{-1}x+\text{cosec}^{-1}x}}{2^{\tan \:\:^{-1}x+\cot \:\:^{-1}x}}\right)$.

 Use ${{\sec }^{-1}}x+\text{cosec}^{-1}x=\dfrac{\pi }{2},{{\tan }^{-1}}x+{{\cot }^{-1}}x=\dfrac{\pi }{2}$ trigonometric identity and then simplify..

  $\text{Consider,} \\  \log\left(\dfrac{2^{\sec \:\:^{-1}x+\text{cosec}^{-1}x}}{2^{\tan \:\:^{-1}x+\cot \:\:^{-1}x}}\right) \\  =\log \:\left(\:\dfrac{2^{\frac{\pi \:}{2}}}{2^{\frac{\pi \:}{2}}}\:\right)\:\:\:\:\:\:\:\:\left[\:\because \:\sec \:^{-1}x+\text{{cosec}}^{-1}x=\dfrac{\pi }{2},\:\tan \:\:^{-1}x+\cot \:\:^{-1}x=\dfrac{\pi \:\:}{2}\right] \\ =\log 1 =0 $

 Write the value of ${{t}^{2}}$ where $t\,=\,\left[ \dfrac{\sin \left( \dfrac{205}{119}\ell  \right)}{1+\tan \left( \dfrac{205}{119}\ell  \right)} \right]$at $\ell =0$.

 Substitute  $\ell ={{0}^{\circ }}$ is the given expression and solve further.

$\begin{align}  & \text{Write the given expression,} \\  & t=\dfrac{\sin \left( \dfrac{205}{119}\ell  \right)}{1+\tan \left( \dfrac{205}{119}\ell  \right)} \\  & \text{Substitute }\ell =0 \\  & t=\dfrac{\sin \left( \dfrac{205}{119}\times {{0}^{\circ }} \right)}{1+\tan \left( \dfrac{205}{119}\times {{0}^{\circ }} \right)} \\  & =\dfrac{\sin 0{}^\circ }{1+\tan 0{}^\circ }=\dfrac{0}{1+0}=0 \\  & t=0 \\  & \Rightarrow {{t}^{2}}=0 \\ \end{align}$

 What is the simplified form of the expression ${{e}^{{{\tan }^{2}}\left( \dfrac{37}{5}\ell  \right)\,+\,{{\cot }^{2}}\left( \dfrac{37}{5}\ell  \right)}}$.

 Add and subtract $2$ in the power and convert the expression of power in whole square form. 

 $\begin{align}  & \text{Consider the given expression} \\  & {{e}^{{{\tan }^{2}}\left( \dfrac{37}{5}\ell  \right)+{{\cot }^{2}}\left( \dfrac{37}{5}\ell  \right)}} \\  & \text{Add and substitute 2 }as\text{ }shown, \\  & {{e}^{{{\tan }^{2}}\left( \dfrac{37}{5}\ell  \right)+{{\cot }^{2}}\left( \dfrac{37}{5}\ell  \right)+2-2}} \\  & ={{e}^{{{\tan }^{2}}\left( \dfrac{37}{5}\ell  \right)+{{\cot }^{2}}\left( \dfrac{37}{5}\ell  \right)+2\tan \left( \dfrac{37}{5}\ell  \right)\cot \left( \dfrac{37}{5}\ell  \right)-2}}\text{  }\left[ \because \tan \theta .\cot \theta =1 \right] \\  & ={{e}^{{{\left[ \tan \left( \dfrac{37}{5}\ell  \right)+\cot \left( \dfrac{37}{5}\ell  \right) \right]}^{2}}-2}} \\ \end{align}$

 Simplify:$\left[ \dfrac{{{\cos }^{2}}\left( \dfrac{2}{3}p \right)+{{\sin }^{2}}\left( \dfrac{2}{3}p \right)}{{{\cos }^{2}}\left( \dfrac{2p}{3} \right)-{{\sin }^{2}}\left( \dfrac{2p}{3} \right)} \right].\,\,\sec \,\left( \dfrac{4p}{3} \right)$.

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

 $\begin{align}  & \text{Write the given expression,} \\  & \left[ \dfrac{{{\cos }^{2}}\left( \dfrac{2p}{3} \right)+{{\sin }^{2}}\left( \dfrac{2p}{3} \right)}{{{\cos }^{2}}\left( \dfrac{2p}{3} \right)-{{\sin }^{2}}\left( \dfrac{2p}{3} \right)} \right].\sec \left( \dfrac{4p}{3} \right) \\  & =\dfrac{1}{{{\cos }^{2}}\left( \dfrac{2p}{3} \right)-{{\sin }^{2}}\left( \dfrac{2p}{3} \right)}.\sec \left( \dfrac{4p}{3} \right)\text{  }\left[ \because {{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1 \right] \\  & =\dfrac{1}{\cos 2\times \left( \dfrac{2p}{3} \right)}.\sec \left( \dfrac{4p}{3} \right)\text{  }\left[ \because {{\cos }^{2}}\theta -{{\sin }^{2}}\theta =\cos 2\theta  \right] \\  & =\dfrac{1}{\cos \left( \dfrac{4p}{3} \right)}.\sec \left( \dfrac{4p}{3} \right) \\  & \Rightarrow {{\sec }^{2}}\left( \dfrac{4p}{3} \right) \\ \end{align}$