NIMCET 2019 Question Paper with solutions

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Question : 21 of 120

Marks: +1, -0

If

\sin^2x

tan x +

\cos^2x

cot x -

2\sin^2x

= 1 + tan x + cot x, x ∊ (0, π), then x

${3π}/12,{5π}/12$
${5π}/12,{7π}/12$
${7π}/12,{11π}/12$
${7π}/12,{9π}/12$

Solution:

We have,

\sin^{2} x\tan x+\cos^{2} x\cot x-\sin 2x=1+\tan x+\cot x, x \in(0, \pi)

⇒1+\tanx(1-\sin^{2} x)+\cot x(1-\cos^{2} x)+\sin 2x=0

⇒1+\tan x\cos^{2} x+\cot x\sin^{2} x+\sin 2x=0

⇒1+\sin x \cos x+\cos x \sin x+\sin 2 x=0

⇒1+2 \sin x \cos x+\sin 2x=0

⇒1+2\sin2x=0

\sin 2x=-\;\;{1}/{2}

\sin 2x=π+\;\;{π}/{6}

or

2π-\;\;{π}/{6}

x=\;\;{7π}/{12}

or

\;\;{11π}/{12}

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