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The Value of Cos ( 90 ° − θ ) Sec ( 90 ° − θ ) Tan θ C O S E C ( 90 ° − θ ) Sin ( 90 ° − θ ) Cot ( 90 ° − θ ) + Tan ( 90 ° − θ ) Cot θ - Mathematics

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Question

The value of

\[\frac{\cos \left( 90°- \theta \right) \sec \left( 90°- \theta \right) \tan \theta}{cosec \left( 90°- \theta \right) \sin \left( 90° - \theta \right) \cot \left( 90°- \theta \right)} + \frac{\tan \left( 90° - \theta \right)}{\cot \theta}\] 

 

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MCQ

Solution

We have to find:  \[\frac{\cos \left( 90°- \theta \right) \sec \left( 90°- \theta \right) \tan \theta}{cosec \left( 90°- \theta \right) \sin \left( 90° - \theta \right) \cot \left( 90°- \theta \right)} + \frac{\tan \left( 90° - \theta \right)}{\cot \theta}\] 

so 

\[\frac{\cos \left( 90°- \theta \right) \sec \left( 90°- \theta \right) \tan \theta}{cosec \left( 90°- \theta \right) \sin \left( 90° - \theta \right) \cot \left( 90°- \theta \right)} + \frac{\tan \left( 90° - \theta \right)}{\cot \theta}\] 

= `(sin θ cosec θ tan θ) /(sec θ  cos θ  tan θ )+cot θ / cot θ ` 

=` (1 xx tan θ) /(1xx tan θ )+cot θ /cot θ ` 

=`1+1` 

=`2` 

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Chapter 10: Trigonometric Ratios - Exercise 10.5 [Page 58]

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RD Sharma Mathematics [English] Class 10
Chapter 10 Trigonometric Ratios
Exercise 10.5 | Q 22 | Page 58
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