Question

answer it

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Answer 1

To prove :

$\dfrac{sin\theta + sin\;2\theta}{1+cos\theta+cos\;2\theta}=tan\theta$

Proof :

Taking LHS

$\implies LHS = \dfrac{sin\theta + sin\;2\theta}{1+cos\theta+cos\;2\theta}$

using trigonometric identity

sin 2θ = 2 sinθ cosθ

$\implies LHS = \dfrac{sin\theta + (2 \;sin\theta\; cos\theta)}{1+cos\theta+cos\;2\theta}$

using trigonometric identity

cos 2θ = 2 cos²θ - 1

$\implies LHS = \dfrac{sin\theta + 2 \;sin\theta\; cos\theta}{1+cos\theta+(2cos^2\theta-1)}$

$\implies LHS = \dfrac{sin\theta + 2 \;sin\theta\; cos\theta}{1+cos\theta+2cos^2\theta-1}$

$\implies LHS = \dfrac{sin\theta + 2 \;sin\theta\; cos\theta}{cos\theta+2cos^2\theta}$

$\implies LHS = \dfrac{sin\theta ( 1 + 2 \;cos\theta)}{cos\theta( 1 + 2cos\theta)}$

1 + 2 cos θ will be cancelled

$\implies LHS = \dfrac{sin\theta}{cos\theta}$

$\implies LHS = tan\theta = RHS$

Hence, proved.

More related trigonometric identities

$\longrightarrow sin(-x) = -sin x$

$\longrightarrow cos (-x) = cos x$

$\longrightarrow cos(x+y) = cos x \; cos y - sinx \;sin y$

$\longrightarrow cos ( x - y ) = cos x \; coy + sinx \; siny$

$\longrightarrow sin(x+y) = sinx\;cosy+cosx\;siny$

$\longrightarrow sin(x-y) = sinx\;cosy - cosx\;siny$

$\longrightarrow tan(x+y) = \dfrac{tanx+tany}{1-tanx\;tany}$

$\longrightarrow tan(x-y) = \dfrac{tanx - tany }{1+tanx\;tany}$

$\longrightarrow cot(x+y) = \dfrac{cotx\;coty-1}{cot y + cot x }$

$\longrightarrow cot (x - y ) = \dfrac{cotx\;coty+1}{coty-cotx}$