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  • The expression  <img alt="\frac{\cos^2\theta(\sec\theta-\tan\theta)}{\sec\theta+tan\theta}" src="http://entrancecorner.oncodecogs.com/gif.latex?%5Cfrac%7B%5Ccos%5E2%5Ctheta%28%5Csec%5Cthe

The expression  \frac{\cos^2\theta(\sec\theta-\tan\theta)}{\sec\theta+tan\theta} can be simplified to 

 

Option: 1

1-\sin\theta


Option: 2

(1-\sin\theta)^2


Option: 3

\sin^2\theta


Option: 4

(1-\cos\theta)^2


Answers (1)

best_answer

Trigonometric Identities

\\\mathrm{\sin^2\mathit{t}+\cos^2\mathit{t}=1}\\\mathrm{1+\tan^2\mathit{t}=\sec^2\mathit{t}}\\\mathrm1+{\cot^2\mathit{t}=\csc^2\mathit{t}}\\\mathrm{\tan \mathit{t}=\frac{\sin \mathit{t}}{\cos \mathit{t}},\;\;\cot \mathit{t}=\frac{\cos\mathit{t}}{\sin\mathit{t}}}

 

Now,

\frac{\cos^2\theta(\sec\theta-tan\theta)}{\sec\theta+tan\theta}=\frac{\cos^2\theta(\sec\theta-tan\theta)}{\sec\theta+tan\theta}\frac{\sec\theta-\tan\theta}{\sec\theta-tan\theta}\\ \\=\frac{(\sec\theta-\tan\theta)^2\cos^2\theta}{\sec^2\theta-tan^2\theta}\\\\ =(\frac{1}{\cos\theta}-\frac{\sin\theta}{\cos\theta})^2\cos^2\theta\\\\ =(1-\sin\theta)^2

Posted by

manish painkra

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