Question

1+cos theta+sin theta/1+costheta-sintheta=1+sintheta/costheta

Answer 1

Answer:

$\frac{1+cos\theta + sin\theta}{1+cos\theta - sin\theta}=\frac{1+sin\theta}{cos\theta}$

L.H.S.

$\frac{1+cos\theta + sin\theta}{1+cos\theta - sin\theta}$

$=\frac{1+cos\theta + sin\theta}{1+cos\theta - sin\theta}\times \frac{1+cos\theta + sin\theta}{1+cos\theta + sin\theta}$

$=\frac{(1+cos\theta +sin\theta)^2}{(1+cos\thta)^2-sin^2\theta}$

$=\frac{(1+cos\theta)^2+sin^2\theta + 2 sin\theta(1+cos\theta)}{1+cos^2\theta + 2cos\theta - sin^2 \theta}$

$=\frac{1+cos^2\theta + 2 cos\theta + sin^2\theta + 2 sin\theta(1+cos\theta)}{1+cos^2\theta + 2cos \theta - 1 + cos^2 \theta}$

$=\frac{1+1+2cos \theta + 2 sin \theta( 1+cos\theta)}{2cos^2\theta + 2 cos\theta}$    $(sin^2\theta + cos^2 \theta = 1)$

$=\frac{2+2cos\theta + 2sin\theta cos\theta + 2 sin \theta}{2cos^2 \theta + 2 cos \theta }$

$=\frac{2(1+cos\theta)+2sin \theta ( 1+ cos \theta)}{2cos \theta (cos \theta + 1)}$

$=\frac{1+sin\theta }{cos \theta }$

R.H.S.

Hence, proved.

Answer 2

Hence proved that

1 + cos θ + sinθ/1 + cosθ - sinθ = 1 + sinθ/cosθ

Explanation:

Given,

L.H.S. = 1 + cos θ + sinθ/1 + cosθ - sinθ

Multiplying the numerator and denominator by (1 + sin θ)

= {(1 + cos θ + sin θ} {(1 + + sin θ)}/{(1 + cos θ - sin θ)} {1 + sin θ}

= {1 + cos θ + sin θ} {(1 + + sin θ)}/{1 + cos θ) - sinθ + sinθ + sinθcosθ - sin^2 θ}

= (1 + cosθ + sinθ) (1 + sin θ)/(cos^2 θ + cosθ + sinθ cos θ)

= (1 + cosθ + sinθ) (1 + sinθ)/ cos θ (cosθ + 1 + sinθ)

= 1 + sinθ/cosθ

= R.H.S.

Learn more: If cos θ+ sin θ=√2 cos θ, show that cos θ-sin θ=√2 sin θ

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