Question

answer as soon as possible
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Answer 1

Step-by-step explanation:

To show : 1 + cotθ.cot2θ = cotθ.cosec2θ.

1 + cotθ.cot2θ = cotθ.cosec2θ

1 = cotθ.cosec2θ - cotθ.cot2θ

1 = cotθ (cosec 2θ - cot 2θ)

1/cotθ = cosec 2θ - cot 2θ

tanθ = 1/sin 2θ - cos 2θ/sin 2θ

sin θ/ cosθ = (1- cos 2θ)/sin 2θ (i)

Now, we know that,

cos (θ+θ) = cos θ.cos θ - sin θ.sin θ

= cos^2 θ - sin^2 θ

and, sin(θ+θ) = sin θ.cos θ + sin θ.cos θ

= 2(sin θ.cos θ);

Cos 2θ = cos (θ+θ)

and, sin2θ = sin (θ+θ)

Thus, Cos 2θ = cos^2 θ - sin^2 θ;

sin2θ = 2(sin θ.cos θ);

Now, put these two values in equation (i);

Thus, sin θ/ cosθ = (1- cos 2θ)/sin 2θ;

sin θ/ cosθ = (1- (cos^2 θ - sin^2 θ)) / 2(sin θ.cos θ);

sin θ/ cosθ = ((cos^2 θ + sin^2 θ)- (cos^2 θ - sin^2 θ)) / 2(sin θ)(cos θ);

sin θ = (cos^2 θ + sin^2 θ - cos^2 θ + sin^2 θ) / 2(sin θ);

sin θ = (2sin^2 θ) / 2(sin θ);

sin θ = (2)(sin θ)(sin θ) / 2(sin θ);

sin θ = sin θ;

Thus, LHS = RHS

That's all.