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

Quotient identities Quiz 2

mathematics

 Find the value of $co{{s}^{4}}\theta +co{{s}^{2}}\theta $ if it is given that $\left( \dfrac{1}{{{\cot }^{2}}\theta }+1 \right)={{\cot }^{2}}\theta $.

 Simplify the given expression in terms of $\sin \theta \text{ and }\cos \theta $ using the identities ${{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta ,\text{ }{{\sin }^{2}}~\theta =1-{{\cos }^{2}}\theta \text{ and }\tan \theta =\dfrac{\sin \theta }{\cos \theta }$.

  $\left( \dfrac{1}{{{\cot }^{2}}\theta }+1 \right)={{\cot }^{2}}\theta \Rightarrow \dfrac{1}{{{\cot }^{2}}\theta }\left( \dfrac{1}{{{\cot }^{2}}\theta }+1 \right)=1 \\  \dfrac{1}{{{\cot }^{4}}\theta }+\dfrac{1}{{{\cot }^{2}}\theta }=1\Rightarrow {{\tan }^{4}}\theta +{{\tan }^{2}}\theta =1 \\  {{\tan }^{2}}\theta ({{\tan }^{2}}\theta +1)\Rightarrow {{\tan }^{2}}\theta {{\sec }^{2}}\theta =1\text{ }\{\because {{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta \} \\  \Rightarrow \dfrac{{{\sin }^{2}}~\theta }{{{\cos }^{4}}~\theta }=1\Rightarrow 1-{{\cos }^{2}}\theta ={{\cos }^{4}}~\theta \text{  }\{\because {{\sin }^{2}}~\theta =1{{\cos }^{2}}\theta \} \\  \Rightarrow {{\cos }^{4}}\theta +{{\cos }^{2}}~\theta =1$

 If $\sec \phi =-2$and$\pi < \phi < \dfrac{3\pi }{2}$, find the value of $4ta{{n}^{2}}\phi - 3cose{{c}^{2}}\phi $

 First find the value of $\cos \phi $ then $\sin \phi $ in order to get the value of the required expression as $\sin \theta =\dfrac{1}{\csc \theta }\text{ and }\tan \theta =\dfrac{\sin \theta }{\cos \theta }$.

  $\sec \phi =-2\Rightarrow \cos \phi =\dfrac{1}{2} \\  \sin \phi =\pm \sqrt{1-co{{s}^{2}}\phi }\Rightarrow \sin \phi =-\sqrt{1-\dfrac{1}{4}}=-\dfrac{\sqrt{3}}{2} \\ \left[ \text{ Since }f\text{ lies in the third quadrant, value of sin}\phi \text{ is -ve}\text{.} \right] \\  \cos ec\phi =-\dfrac{2}{\sqrt{3}}\text{and }\tan \phi =\dfrac{-\sin \phi }{\cos \phi }=-\sqrt{3} \\ \left[ \text{ Since }f\text{ lies in the third quadrant, value of tan}\phi \text{ is +ve}\text{.} \right] \\  \text{Now, }4{{\tan }^{2}}\phi - 3\cos e{{c}^{2}}\phi =4\times 3 - 3\times \dfrac{4}{3}=8$

 If $7{{\tan }^{2}}\phi =4{{\sec }^{2}}\phi -3$ and $0\le \phi \le \dfrac{\pi }{2}$, then the value of $tan\phi $ is:

 First simplify the given expression in terms of $\sin \phi \text{ and }\cos \phi $ using the identities ${{\sin }^{2}}~\theta =1-{{\cos }^{2}}\theta \text{ and }\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ then put the value of the trigonometric ratio.

$ 7{{\tan }^{2}}\phi =4{{\sec }^{2}}\phi -3\Rightarrow 7\left( \dfrac{si{{n}^{2}}~\phi }{co{{s}^{2}}\phi } \right)=\dfrac{4}{co{{s}^{2}}\phi }-3 \\  7si{{n}^{2}}~\phi +3co{{s}^{2}}\phi =4\Rightarrow 7\text{ }si{{n}^{2}}\phi +3(1-si{{n}^{2}}\phi )=4 \\  \Rightarrow si{{n}^{2}}\phi =\dfrac{1}{4}\Rightarrow sin\phi =\dfrac{1}{2} \\  \therefore cos\phi =\sqrt{(1-si{{n}^{2}}\phi )}=\sqrt{1-\dfrac{1}{4}}=\dfrac{\sqrt{3}}{2} \\  \therefore tan\phi =\dfrac{sin\phi }{cos\phi }\text{=}\dfrac{1}{2}\times \dfrac{2}{\sqrt{3}}=\dfrac{1}{\sqrt{3}}$

 For any real values of $\sqrt{\dfrac{\left( \sec \theta - 1 \right)}{\left( \sec \theta +1 \right)}}$, which of the following is true?

 First rationalize the numerator and denominator of the given expression, then use the identities ${{\sec }^{2}}~\theta - 1={{\tan }^{2}}\theta ,\text{ }\dfrac{1}{\sin \theta }=\csc \theta \text{ and }\dfrac{1}{\tan \theta }=\cot \theta $.

 $\sqrt{\dfrac{\left( \sec \theta - 1 \right)}{\left( \sec \theta +1 \right)}}=\sqrt{\dfrac{\left( \sec \theta - 1 \right)}{\left( \sec \theta +1 \right)}}\times \sqrt{\dfrac{\left( \sec \theta - 1 \right)}{\left( \sec \theta -1 \right)}}\text{ }\left[ \text{Rationalizing the numerator and the denominator} \right] \\  \Rightarrow \sqrt{\dfrac{{{\left( \sec \theta - 1 \right)}^{2}}}{\left( {{\sec }^{2}}\theta -1 \right)}}=\sqrt{\dfrac{{{\left( \sec \theta - 1 \right)}^{2}}}{{{\tan }^{2}}\theta }}\text{      }[\because \text{ }{{\sec }^{2}}~\theta - 1={{\tan }^{2}}\theta ] \\  \Rightarrow \dfrac{\left( \sec \theta - 1 \right)}{\tan \theta }=\dfrac{\sec \theta }{\tan \theta }-\dfrac{1}{\tan \theta }=\dfrac{1}{\sin \theta }-\dfrac{1}{\tan \theta }\text{ } \\  \Rightarrow \csc \theta - \cot \theta \text{    }\left[ \because \text{ }\dfrac{1}{\sin \theta }=\csc \theta \text{ and }\dfrac{1}{\tan \theta }=\cot \theta  \right]$

 If ${{\tan }^{2}}\theta - 1=\alpha \tan \theta $ and $\cos 2\theta =\beta \left( \cos \theta +\sin \theta  \right)$, then the value of $\left( {{\alpha }^{2}}+4 \right)\left( {{\beta }^{2}}-1 \right)$ is

 Divide the first given expression by $\tan \theta $ on both the sides and then simplify both the given expression in terms of $\sin \theta \text{ and }\cos \theta $ using the identities ${{\sin }^{2}}~\theta =1-{{\cos }^{2}}\theta \text{ and }\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ . Now, multiply both for the required RHS.

 $\text{We have, }{{\tan }^{2}}\theta - 1=\alpha \tan \theta \Rightarrow \tan \theta - \dfrac{1}{\tan \theta }=\alpha \Rightarrow \tan \theta - \cot \theta =\alpha  \\  \Rightarrow {{\tan }^{2}}\theta +{{\cot }^{2}}~\theta - 2\tan \theta \cot \theta ={{\alpha }^{2}}\Rightarrow \dfrac{{{\sin }^{2}}~\theta }{{{\cos }^{2}}~\theta }+\dfrac{{{\cos }^{2}}~\theta }{{{\sin }^{2}}~\theta }={{\alpha }^{2}}\text{+ }2 \\  \Rightarrow {{\sin }^{4}}\theta +{{\cos }^{4}}\theta =({{\alpha }^{2}}~+\text{ }2){{\sin }^{2}}~\theta {{\cos }^{2}}~\theta \Rightarrow 1-2{{\sin }^{2}}~\theta {{\cos }^{2}}\theta =\left( {{\alpha }^{2}}~+\text{ }2 \right){{\sin }^{2}}~\theta co{{s}^{2}}~\theta  \\  \therefore {{\alpha }^{2}}~+\text{ }4=\dfrac{1}{{{\sin }^{2}}~\theta \text{ }{{\cos }^{2}}~\theta } \\ \text{And, }\cos 2\theta =\beta \left( \cos \theta +\sin \theta  \right)\Rightarrow {{\cos }^{2}}\theta - {{\sin }^{2}}\theta =\beta \left( \cos \theta +\sin \theta  \right)\Rightarrow \cos \theta -\sin \theta =\beta  \\  \cos \theta- \sin \theta =\beta {{\left( \cos \theta - \sin \theta  \right)}^{2}}~={{\beta }^{2}}\Rightarrow {{\beta }^{2}}-1=2\sin \theta  \cos \theta \Rightarrow {{\left( \beta - 1 \right)}^{2}}~=4{{\sin }^{2}}~\theta {{\cos }^{2}}\theta  \\  \therefore \left( {{\alpha }^{2}}~+\text{ }4 \right){{\left( \beta - 1 \right)}^{2}}=4$