NIMCET 2019 — Mathematics PYQ

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Question

NIMCET 2019 — Mathematics PYQ

NIMCET | Mathematics | 2019

If $sin^{2} x tan x + cos^{2} cot x - sin 2 x = 1 + tan x + cot x$ , $x \in (0, π)$ , then

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Explanation

Concept:
- $tan θ = \frac{s i n θ}{c o s θ}$
- $sin^{2} x + cos^{2} x = 1$
- $sin^{4} x + cos^{4} x = 1 - 2 sin^{2} x cos^{2} x$

Calculation:

Given: $sin^{2} x tan x + cos^{2} cot x - sin 2 x = 1 + tan x + cot x$

$\Rightarrow \frac{s i n ^{2} x}{c o s x} + \frac{c o s ^{3} x}{s i n x} - sin 2 x = 1 + \frac{s i n x}{c o s x} + \frac{c o s x}{s i n x}$

$\Rightarrow \frac{s i n ^{2} x + c o s ^{4} x}{s i n x c o s x} - sin 2 x = 1 + \frac{s i n ^{2} x + c o s ^{2} x}{s i n x c o s x}$

$\Rightarrow \frac{1 - 2 s i n ^{2} x c o s ^{2} x}{s i n x c o s x} - sin 2 x = 1 + \frac{1}{s i n x c o s x}$

$\Rightarrow \frac{1}{s i n x c o s x} - \frac{2 s i n ^{2} x c o s ^{2} x}{s i n x c o s x} - sin 2 x = 1 + \frac{1}{s i n x c o s x}$

$\Rightarrow - 2 sin x cos x - sin 2 x = 1$

$\Rightarrow - 2 sin 2 x = 1$

$\Rightarrow sin 2 x = - \frac{1}{2}$

$\Rightarrow 2 x = \frac{7 π}{6}, \frac{11 π}{6}$

$\Rightarrow x = \frac{7 π}{12}, \frac{11 π}{12}$

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