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Prove that coseccot(180∘+θ)sin(90∘-θ)cos(-θ)sin(270∘+θ)tan(-θ)cosec(360∘+θ) = cos2θ cotθ

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प्रश्न

Prove that `(cot(180^circ + theta) sin(90^circ - theta) cos(- theta))/(sin(270^circ + theta) tan(- theta) "cosec"(360^circ + theta))` = cos2θ cotθ

योग
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उत्तर

`(cot(180^circ + theta) sin(90^circ - theta) * cos(- theta))/(sin(270^circ + theta) tan(- theta)  "cosec"(360^circ + theta))` 

= `(cot theta* costheta costheta)/(- cos theta xx - tantheta xx "cosec" theta)`

= `(cot theta * cos^2theta)/(cos theta tan theta  "cosec" theta)`

= `(cot theta * cos^2theta)/(cos theta * sintheta/costheta * 1/sin theta)`

= `cos^2theta * cottheta`

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Trigonometric Functions and Their Properties
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 3: Trigonometry - Exercise 3.3 [पृष्ठ १०४]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
अध्याय 3 Trigonometry
Exercise 3.3 | Q 4 | पृष्ठ १०४

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