Question

1/2{sinteta/1+costeta+1+costeta/sinteta}=1/sinteta

Answer 1

Answer:

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Answer 2

We have to prove the following equation.

$\quad \displaystyle \bf \frac{1}{2} \bigg\{\frac{sin\:\theta}{1 + cos\:\theta}+\frac{1 + cos\:\theta}{sin\:\theta}\bigg\} = \frac{1}{sin\theta}$

$\sf Solving \: L.H.S \: separately$

$= \quad \displaystyle \bf \frac{1}{2}\bigg\{\frac{sin \: \theta}{1 + cos \: \theta} + \frac{1 + cos \: \theta}{sin \: \theta} \bigg\}$

$= \quad \displaystyle \bf \frac{1}{2}\bigg\{ \frac{ {(sin \: \theta)}^{2} + {(1 + cos \: \theta)}^{2} }{sin \: \theta(1 + cos \: \theta)}\bigg\}$

$= \displaystyle \bf \quad \frac{1}{2} \bigg\{ \frac{ {sin}^{2} \theta + {1}^{2} + {cos}^{2}\theta + 2cos \theta }{sin \theta(1 + cos \theta)} \bigg\}$

$= \displaystyle \quad \bf \frac{2 + 2cos \theta}{2sin \theta(1 + cos \theta)} \qquad \bigg( \because {sin}^{2} \theta + {cos}^{2} \theta = 1 \bigg)$

$= \displaystyle \quad \bf \frac{ \not{2} \cancel{(1 + cos \theta)}}{ \not{2}sin \theta \cancel{(1 + cos \theta)}}$

$= \displaystyle \quad \bf \frac{1}{sin \theta}$

Since, L.H.S = R.H.S

Hence, Proved.

Formulae are must, So better learn and remember them. If you get in any other question like this, Try converting terms into sine and cosine.

Some Formulae :-

◉ sin² θ + cos² θ = 1

◉ 1 + tan² θ = sec² θ

◉ 1 + cot² θ = cosec² θ

◉ sin (A + B) = sinAcosB + cosAsinB

◉ sin 2θ = 2sinθcosθ

◉ 1 + cos θ = 2 cos² θ/2

◉ 1 - cos θ = 2 sin² θ/2