Evaluatesin180-x-270-x-tan270-x-180-x

We have, sin(180^∘ x).(270^∘+x).tan(270^∘ x).(180^∘+x) (∵ sin(180^∘ x)=sin x , (270^∘+x)=( tan x) and tan(270^∘ x)= x, (180^∘+x)= x) ⇒ sin x.( tan x). x.( x) ...

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Byju's Answer

Standard XII

Physics

Introduction

Evaluatesin18...

Question

Evaluate
$sin (180^{\circ} - x) . cot (270^{\circ} + x) . tan (270^{\circ} - x) . sec (180^{\circ} + x)$

sec x

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cot x

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sin x

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tan x

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Solution

The correct option is D $tan x$
We have, $sin (180^{\circ} - x) . cot (270^{\circ} + x) . tan (270^{\circ} - x) . sec (180^{\circ} + x)$
$(∵ sin (180^{\circ} - x) = sin x, cot (270^{\circ} + x) = (- tan x) and tan (270^{\circ} - x) = cot x, sec (180^{\circ} + x) = - sec x)$
$\Rightarrow sin x . (- tan x) . cot x . (- sec x)$
$(∵ tan x . cot x = 1)$
$\Rightarrow sin x . sec x = tan x$

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Evaluate sin(180∘−x).cot(270∘+x).tan(270∘−x).sec(180∘+x)

The correct option is D tanxWe have, sin(180∘−x).cot(270∘+x).tan(270∘−x).sec(180∘+x) (∵ sin(180∘−x)=sinx, cot(270∘+x)=(−tanx)andtan(270∘−x)=cotx,sec(180∘+x)=−secx) ⇒ sinx.(−tanx).cotx.(−secx) (∵tanx .cotx=1) ⇒ sinx .secx=tanx

Evaluate
$sin (180^{\circ} - x) . cot (270^{\circ} + x) . tan (270^{\circ} - x) . sec (180^{\circ} + x)$