Question

plz answer fast.........​

Answer 1

Given :

• An equation, $\bf \sqrt{ \dfrac{1 + cos \: \theta }{1 - cos \: \theta} } = cosec \: \theta + cot \: \theta$

To Prove :

• The given equation, $\bf \sqrt{ \dfrac{1 + cos \: \theta }{1 - cos \: \theta} } = cosec \: \theta + cot \: \theta$

Proof :

Given equation,

$\bf \sqrt{ \dfrac{1 + cos \: \theta }{1 - cos \: \theta} } = cosec \: \theta + cot \: \theta$

Solve the equation of L.H.S. (Left hand side),

$\bf \implies \: L.H.S. = \sqrt{ \dfrac{1 + cos \: \theta}{1 - cos \: \theta} } \\ \\ \\ \bf \implies \sqrt{ \dfrac{1 + cos \: \theta}{1 - cos \: \theta} \times \dfrac{1 + cos \: \theta}{1 + cos \: \theta} }$

Using identity,

• (a - b) (a + b) = a² - b²

$\bf \implies \sqrt{ \dfrac{ {(1 + cos \: \theta)}^{2} }{ {(1)}^{2} - {(cos \: \theta)}^{2} } } \\ \\ \\ \bf \implies \sqrt{ \dfrac{{(1 + cos \: \theta)}^{2} }{1 - {cos}^{2} \: \theta} }$

We know that,

$\bullet \: \: \: \bf{1 - {cos}^{2} \: \theta} = {sin}^{2} \: \theta$

$\bf \implies \sqrt{\dfrac{ {(1 + cos \: \theta)}^{2} }{ {sin}^{2} \: \theta} } \\ \\ \\ \bf \implies \frac{1 + cos \: \theta}{sin \: \theta} \\ \\ \\ \bf \implies \dfrac{1}{sin \: \theta} + \dfrac{cos \: \theta}{sin \: \theta}$

We know that,

$\bullet \: \: \: \bf \dfrac{1}{sin \: \theta} = cosec \: \theta \\ \\ \\ \bullet \: \: \: \bf \dfrac{cos \: \theta}{sin \: \theta} = cot \: \theta$

So,

$\bf \implies cosec \: \theta + cot \: \theta \\ \\ \bf\bf \implies \: R.H.S.$

Hence Proved !

Answer 2

Given:-

• $\sf\ \sqrt{\dfrac{1+cos θ}{1-cosθ}} = cosecθ+cotθ$

To Find:-

• An equation,

$\sf\ \sqrt{\dfrac{1+cos θ}{1-cosθ}} =cosecθ+cotθ$

Solution:-

$\sf\ LHS :- \sqrt{\dfrac{1+cos θ}{1-cosθ}} =cosecθ+cotθ$

$\sf\ Multiplying\: numerator\: and \: denominator$

$\sf\ with\:1+cosθ\: we\: get$

$\Longrightarrow\: \sf\ \sqrt{\dfrac{1+cos θ}{1-cosθ} × \dfrac{1+cosθ}{1+cosθ}}$

$\Longrightarrow\: \sf\ \sqrt{ \dfrac{(1+cos θ)^2} {1-cos^2θ} }$

$\Longrightarrow\: \sf\ \dfrac{1+cos θ}{sinθ}$

$\Longrightarrow\: \sf\ \dfrac{1}{sinθ} + \dfrac{cosθ}{sinθ}$

$\Longrightarrow\: \sf\ cosecθ + cotθ$

$\Longrightarrow\: \sf\ RHS$

Hence proved!