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This is one of our many quizzes on Proving trigonometric identities. Click on this to access more such competitive quizzes on Trigonometry.

Proving trigonometric identities Quiz 2

mathematics

 Which of the following is equivalent to $\sec \left( \dfrac{4}{5} \right)+\tan \left( \dfrac{4}{5} \right)$

 First use $\sec \theta =\dfrac{1}{\cos \theta }$ and $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and then use complementary angles of $\sin \theta $ and $\cos \theta $.

  $\text{Consider, }\sec \left( \dfrac{4}{5} \right)+\tan \left( \dfrac{4}{5} \right) \\ \Rightarrow \dfrac{1}{\cos \left( \dfrac{4}{5} \right)}+\dfrac{\sin \left( \dfrac{4}{5} \right)}{\cos \left( \dfrac{4}{5} \right)} \\ \Rightarrow \dfrac{1+\sin \left( \dfrac{4}{5} \right)}{\cos \left( \dfrac{4}{5} \right)} \\ \Rightarrow \dfrac{1+\cos \left( \dfrac{\pi }{2}-\dfrac{4}{5} \right)}{\sin \left( \dfrac{\pi }{2}-\dfrac{4}{5} \right)}\left[ \begin{align}  & \because \sin \theta =\cos \left( \dfrac{\pi }{2}-\theta  \right) \\  & \cos \theta =\sin \left( \dfrac{\pi }{2}-\theta  \right) \\ \end{align} \right]$

 Evaluate ${{\left( {{\sin }^{2}}\dfrac{2m}{3}+{{\cos }^{2}}\dfrac{2m}{3}+{{\tan }^{2}}\dfrac{2m}{3} \right)}^{2}}$ and write the value.

 Use ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$. Then use the identity ${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1$ to solve.

  ${{\left[ {{\sin }^{2}}\left( \dfrac{2m}{3} \right)+{{\cos }^{2}}\left( \dfrac{2m}{3}\, \right)+{{\tan }^{2}}\left( \dfrac{2m}{3}\, \right) \right]}^{2}} \\ \Rightarrow {{\left[ 1+{{\tan }^{2}}\dfrac{2m}{3} \right]}^{2}}={{\left[ {{\sec }^{2}}\left( \dfrac{2m}{3} \right) \right]}^{2}} \\ \Rightarrow {{\sec }^{4}}\left( \dfrac{2m}{3} \right)$

 Write the value of ${{\left( 3+4\sin \dfrac{y}{2} \right)}^{2}}+{{\left( 3+4\cos \dfrac{y}{2} \right)}^{2}}$

 First solve the given expression algebraically and then use ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$.

  $\text{Consider the expression.} \\ {{\left[ 3+4\sin \left( \dfrac{y}{2} \right) \right]}^{2}}+{{\left[ 3+4\cos \left( \dfrac{y}{2} \right) \right]}^{2}} \\ \text{Use }{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}} \\ \Rightarrow 9+24\sin \left( \dfrac{y}{2} \right)+16{{\sin }^{2}}\left( \dfrac{y}{2} \right)+9+24\cos \left( \dfrac{y}{2} \right)+16{{\cos }^{2}}\left( \dfrac{y}{2} \right) \\ \Rightarrow 18+24\left( \sin \dfrac{y}{2}+\cos \dfrac{y}{2} \right)+16\left( {{\sin }^{2}}\dfrac{y}{2}+{{\cos }^{2}}\dfrac{y}{2} \right) \\ \Rightarrow 18+24\left( \sin \dfrac{y}{2}+\cos \dfrac{y}{2} \right)+16 \\ \Rightarrow 34+24\left( \sin \dfrac{y}{2}+\cos \dfrac{y}{2} \right)$

 Write the equivalent value of the trigonometric expression $\dfrac{3\sin \dfrac{g}{2}+\cos \dfrac{g}{2}}{3\sin \dfrac{g}{2}-\cos \dfrac{g}{2}}$

 After dividing numerator and denominator by $\sin \left( \dfrac{g}{2} \right)$, use $\dfrac{\cos \theta }{\sin \theta }=\cot \theta $ to solve further.

  $\text{Consider the expression,} \\ \dfrac{3\sin \dfrac{g}{2}+\cos \dfrac{g}{2}}{3\sin \dfrac{g}{2}-\cos \dfrac{g}{2}} \\ \text{Divide numerator and denominator by }\sin \left( \dfrac{g}{2} \right) \\ \Rightarrow \dfrac{\dfrac{3\sin \dfrac{g}{2}}{\sin \dfrac{g}{2}}+\dfrac{\cos \dfrac{g}{2}}{\sin \dfrac{g}{2}}}{\dfrac{3\sin \dfrac{g}{2}}{\sin \dfrac{g}{2}}-\dfrac{\cos \dfrac{g}{2}}{\sin \dfrac{g}{2}}} \\ \Rightarrow \dfrac{3+\cot \dfrac{g}{2}}{3-\cot \dfrac{g}{2}}$

 Simply by the trigonometric expression ${{\left( \dfrac{7-7\cos 5\theta }{7+7\cos 5\theta } \right)}^{\frac{3}{4}}}.$

 Take $7$ common and then use half angle formulae to get the desired result.

  $\text{Write the trigonometric expression.} \\ {{\left( \dfrac{7-7\cos 5\theta }{7+7\cos 5\theta } \right)}^{\frac{3}{4}}} \\ \Rightarrow {{\left[ \dfrac{7\left( 1-\cos 5\theta  \right)}{7\left( 1+\cos 5\theta  \right)} \right]}^{\frac{3}{4}}} \\ \Rightarrow {{\left( \dfrac{1-\cos 5\theta }{1+\cos 5\theta } \right)}^{\frac{3}{4}}} \\ \Rightarrow {{\left[ \dfrac{2{{\sin }^{2}}\left( \dfrac{5\theta }{2} \right)}{2{{\cos }^{2}}\left( \dfrac{5\theta }{2} \right)} \right]}^{\frac{3}{4}}}\left[ \because 1-\cos x=2{{\sin }^{2}}\dfrac{x}{2},1+\cos x=2{{\cos }^{2}}\dfrac{x}{2} \right] \\ \Rightarrow {{\left( {{\tan }^{2}}\dfrac{5\theta }{2} \right)}^{\frac{3}{4}}}={{\tan }^{\frac{3}{2}}}\left( \dfrac{5\theta }{2} \right)$