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

Cofunction identities Quiz 2

mathematics

 Evaluate: $\dfrac{1}{\sin (75^\circ +\theta )}-\dfrac{1}{\cos (15^\circ -\theta )}$ 

 First write the reciprocal trigonometric functions of both the terms then use the identity $\sec (90^\circ -\theta )= \text{cosec}\theta $ 

  $\dfrac{1}{\sin (75^\circ +\theta )}-\dfrac{1}{\cos (15^\circ -\theta )} \\ \Rightarrow \text{cosec}(75^\circ +\theta )-\sec (15^\circ -\theta ) \\ \Rightarrow \text{cosec}(75^\circ +\theta )-\sec [(90^\circ -(75^\circ +\theta )] \\ \Rightarrow \text{cosec}(75^\circ +\theta )-\text{cosec}(75^\circ +\theta )=0 $

 Find the value of $\tan 5\theta $, given that the ratio between $\tan 8\theta $ and $\tan 2\theta $ is $1$, where $8\theta $ is an acute angle.

 First write the whole trigonometric expression in terms of tan by using the identity $\cot (90^\circ -\theta )=\tan \theta $ then evaluate both sides.

  $\dfrac{\tan 8\theta }{\tan 2\theta }=1 \\ \tan 8\theta =\cot 2 \theta  \\  \tan 8\theta =\tan \left( 90^\circ - 2\theta  \right)  \\   8\theta =90^\circ - 2\theta   \\   10\theta = 90^\circ   \\   \tan 5\theta =\tan 45^\circ =1 $

 Evaluate: $\dfrac{1}{\sin (75^\circ +\theta )}-\dfrac{1}{\cos (15^\circ -\theta )}-\dfrac{1}{\cot (55^\circ +\theta )}+\dfrac{1}{\tan (35^\circ -\theta )}$ 

 First write the reciprocal trigonometric functions of both the terms then use the identity $\sec (90^\circ -\theta )=\text{cosec}\theta $and $\tan (90^\circ -\theta )=\cot \theta $.

  $\dfrac{1}{\sin (75^\circ +\theta )}-\dfrac{1}{\cos (15^\circ -\theta )}-\dfrac{1}{\cot (55^\circ +\theta )}+\dfrac{1}{\tan (35^\circ -\theta )} \\ \Rightarrow \text{cosec}\left( 75^\circ +\theta  \right)\sec \left( 15^\circ - \theta  \right)\tan \left( 55^\circ +\theta  \right)+\cot \left( 35^\circ - \theta  \right)  \\ \Rightarrow \text{cosec}\left( 75^\circ +\theta  \right)\sec \left[ 90^\circ - \left( 75^\circ - \theta  \right) \right]\tan \left( 55^\circ +\theta  \right)+\tan \left[ 90^\circ - \left( 55^\circ \theta - \right) \right]  \\ \Rightarrow \text{cosec}\left( 75^\circ +\theta  \right)\text{cosec}\left( 75^\circ +\theta  \right)\tan \left( 55^\circ +\theta  \right)+\tan \left( 55^\circ +\theta  \right)=0 $

 Find the value of $\dfrac{1}{\cos 21^\circ }-\dfrac{1}{\text{cosec}69^\circ }$in terms of $''x''\text{ and }''y''$ if it is given that $\dfrac{1}{\text{cosec}21^\circ }=\dfrac{x}{y}$.

 First simplify the given relation to get the value of $\sec 21^\circ $ in terms of $''x''\text{ and }''y''$then use $\sin (90^\circ -\theta )=\cos \theta $ in the expression.

  $\text{Here, }\dfrac{1}{\text{cosec}21^\circ }=\dfrac{x}{y} \\ \sin 21^\circ =\dfrac{x}{y} \\    \cos 21^\circ =\sqrt{\left( 1-{{\sin }^{2}}21^\circ  \right)}  \\ \Rightarrow \sqrt{\left\{ 1-\left( \dfrac{{{x}^{2}}}{{{y}^{2}}} \right) \right\}}=\dfrac{\sqrt{\left( {{y}^{2}}-{{x}^{2}} \right)}}{y}  \\   \therefore \sec 21^\circ =\dfrac{y}{\sqrt{\left( {{y}^{2}}-{{x}^{2}} \right)}}  \\   \text{Now, }\dfrac{1}{\cos 21^\circ }-\dfrac{1}{\text{cosec}69^\circ }=\sec 21^\circ - \sin 69^\circ   \\ \Rightarrow \sec 21^\circ - \sin \left( 90^\circ - 21^\circ  \right)  \\ \Rightarrow \sec 21^\circ - \cos 21^\circ   \\ \Rightarrow ~\dfrac{y}{\sqrt{\left( {{y}^{2}} - {{x}^{2}} \right)}}\dfrac{\sqrt{\left( {{y}^{2}} - {{x}^{2}} \right)}}{y}  \\ \Rightarrow \dfrac{{{y}^{2}}\left( {{y}^{2}} - {{x}^{2}} \right)}{y\sqrt{\left( {{y}^{2}} - {{x}^{2}} \right)}}  \\ \Rightarrow \dfrac{{{x}^{2}}}{y\sqrt{\left( {{y}^{2}}-{{x}^{2}} \right)}}$

 Evaluate: ${{\sin }^{2}}{{51}^{\circ }}+\dfrac{1}{{{\sin }^{2}}{{39}^{\circ }}}+\dfrac{1}{{{\sec }^{2}}{{51}^{\circ }}}-{{\csc }^{2}}{{51}^{\circ }}{{\cos }^{2}}{{39}^{\circ }}$ where $\text{cosec}39^\circ =x$ 

 First simplify the given relation then use $\sin (90^\circ -\theta )=\cos \theta $ and $\tan (90^\circ -\theta )=\cot \theta $ in the expression.

  ${{\sin }^{2}}{{51}^{\circ }}+\dfrac{1}{{{\sin }^{2}}{{39}^{\circ }}}+\dfrac{1}{{{\sec }^{2}}{{51}^{\circ }}}-{{\csc }^{2}}{{51}^{\circ }}{{\cos }^{2}}{{39}^{\circ }} \\ \Rightarrow {{{\sin }^{2}}{{51}^{\circ }}}+{{\csc }^{2}}{{39}^{\circ }}+{{\cos}^{2}}{{51}^{\circ }}-\dfrac{1}{{{\sin }^{2}}{{51}^{\circ }}{{\sec }^{2}}{{39}^{\circ }}}  \\  \Rightarrow {{{\sin }^{2}}{{51}^{\circ }}}+{{\csc }^{2}}{{39}^{\circ }}+{{\cos}^{2}}{{51}^{\circ }}-\dfrac{1}{{{\sin }^{2}}\left( {{90}^{\circ }}-{{39}^{\circ }} \right){{\sec }^{2}}{{39}^{\circ }}}  \\  \Rightarrow 1+{{\text{cosec} }^{2}}{{39}^{\circ }}-1 \\ \Rightarrow {{\text{cosec} }^{2}}{{39}^{\circ }}={{x}^{2}}$