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Tan θ Sec θ − 1 + Tan θ Sec θ + 1 is Equal to

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Question

\[\frac{\tan \theta}{\sec \theta - 1} + \frac{\tan \theta}{\sec \theta + 1}\] is equal to 

 

 

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Solution

The given expression is `cot θ/(cotθ-cot 3θ)+tanθ/(tanθ-tan3θ)`

Simplifying the given expression, we have

`cotθ/(cotθ-cot3θ)+ tanθ/(tanθ-tan3θ)` 

= `(cosθ/sin)/(cosθ/sinθ-(cos3θ)/(sin3θ))+(sinθ/cosθ)/(sinθ/sinθ-(sin3θ)/(cos3θ))`

=` (cosθ/sinθ)/((cosθsin 3θ-cos3θsinθ)/(sinθ sin3θ))+ (sin θ/cos θ)/((sinθ cos3θ-sin3θ cosθ)/(cosθ cos3θ))`

=` (cosθ sin3θ)/(cosθ sin3θ-cos3θsinθ)+(sinθ cos3θ)/(sinθ cos3θ-sin3θ sinθ)`

=`(cosθ sin3θ)/(cosθsinθ-cos3θsinθ)+(cos3θ sinθ)/(-1(cosθ sin3θ-cos3θ sinθ))`  

`= (cosθ sin3θ)/(cosθ sin3θ-cos3θsinθ)-(cos3θsinθ)/(cosθsin3θ-cos3θsinθ)` 

`=(cosθsin3θ-cos3θsinθ)/(cosθsin3θ-cos3θsinθ)` 

=1

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Chapter 11: Trigonometric Identities - Exercise 11.4 [Page 57]

APPEARS IN

R.D. Sharma Mathematics [English] Class 10
Chapter 11 Trigonometric Identities
Exercise 11.4 | Q 14 | Page 57

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