Question

tan2x - sin2x = tan2x sin2x


5)1)

Answer 1

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Answer 2

$\huge\underline\mathbb{\red Q\pink{U}\purple{ES} \blue{T} \orange{IO}\green{N :}}$

Prove that : tan² x - sin² x = tan²x sin² x

$\huge\underline\mathbb{\red S\pink{O}\purple{LU} \blue{T} \orange{IO}\green{N :}}$

We should prove that LHS = RHS.

Take LHS = tan² x - sin² x

$\sf\:⟹ tan^{2} \: x - sin^{2} \: x$

• $\bf\: [ ↪ tan^{2} x = \frac{sin^{2} x}{cos^{2} x} ]$

$\sf\:⟹ \frac{sin^{2}x }{cos^{2} x} - sin^{2}$

$\sf\:⟹ sin^{2}x [ \frac{1}{cos^{2}x} - 1 ]$

$\sf\:⟹ sin^{2}x [ \frac{1 - cos^{2}x}{cos^{2}x} ]$

• $\bf\: [↪ 1 - cos^{2} x = sin^{2}x ]$

$\sf\:⟹ sin^{2} x [\frac {sin^{2}x}{cos^{2}x}]$

• $\bf\: [↪ \frac{sin^{2}x}{cos^{2}x} = tan^{2} x ]$

$\sf\:⟹ sin^{2}x \: tan^{2} x$

Hence, LHS = RHS

# IT IS PROVED...

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★ Important identities :

$\bf\:◼ \: sin^{2} \: θ + cos^{2} \: θ = 1$

$\bf\:◼ \: sec^{2} \: θ - tan^{2} \: θ = 1$

$\bf\:◼ \: cosec^{2} \: θ - cot^{2} \: θ = 1$

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