Question

Hola Brainlians. Here's a question to check your Trigonometric skill. ICSE Board : Inverse Trigonometry, Class 10. Give answer fast and no spam.

Proof the given identity.

√[(1 + sin x)/(1 - sin x)] = sec A + tan A

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Answer 1

Answer:

answer in the attachment.

Answer 2

★ Concept :-

Here the concept of Trigonometric Identities have been used. We see that we are given an equation to prove. Most of the students get confused in finding these only. The qúestion seems tough but it is very simple. Firstly we will rationalise the fraction in L.H.S. so we can remove square root. Then we can factorise it using different trignometric identities to find the answer.

Let's do it !!

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★ Solution :-

Given to prove,

$\;\;\bf{\mapsto\;\;\red{\sqrt{\dfrac{1\:+\:\sin x}{1\:-\:\sin x}}\;=\;\sec x\;+\;\tan x}}$

This is the appropriate qúestion.

From here we will get,

$\;\;\sf{\odot\;\;L.H.S\:=\:\green{\sqrt{\dfrac{1\:+\:\sin x}{1\:-\:\sin x}}}}$

And,

$\;\;\sf{\odot\;\;\blue{R.H.S.\;=\;\sec x\;+\;\tan x}}$

Now let's start solving L.H.S. part firstly. So we get,

$\;\;\tt{\rightarrow\;\;L.H.S\:=\:\sqrt{\dfrac{1\:+\:\sin x}{1\:-\:\sin x}}}$

Let's multiply both numerator and denominator of this fraction by (1 + sin x) to rationalise it.

$\;\;\tt{\rightarrow\;\;L.H.S\:=\:\sqrt{\bigg(\dfrac{1\:+\:\sin x}{1\:-\:\sin x}\bigg)\:\times\:\bigg(\dfrac{1\:+\:\sin x}{1\:+\:\sin x}\bigg)}}$

Since multiplying with same terms both numerator and denominator won't cause any change in initial fraction.

$\;\;\tt{\rightarrow\;\;L.H.S\:=\:\sqrt{\dfrac{(1\:+\:\sin x)(1\:+\:\sin x)}{(1\:-\:\sin x)(1\:+\:\sin x)}}}$

The denominator gives us the form of an identity. That is :: (a + b)(a - b) = a² - b²

• Here a = 1

• Here b = sin x

By applying this, we get

$\;\;\tt{\rightarrow\;\;L.H.S\:=\:\sqrt{\dfrac{(1\:+\:\sin x)(1\:+\:\sin x)}{(1)^{2}\:-\:(\sin x)^{2}}}}$

$\;\;\tt{\rightarrow\;\;L.H.S\:=\:\sqrt{\dfrac{(1\:+\:\sin x)(1\:+\:\sin x)}{1^{2}\:-\:\sin^{2}x}}}$

We know that, cos² A = 1 - sin² A which gives us an identity.

• Here A = x

By applying this, we get

$\;\;\tt{\rightarrow\;\;L.H.S\:=\:\sqrt{\dfrac{(1\:+\:\sin x)(1\:+\:\sin x)}{\cos^{2}x}}}$

The numerator can be written as (by multiplying),

$\;\;\tt{\rightarrow\;\;L.H.S\:=\:\sqrt{\dfrac{(1\:+\:\sin x)^{2}}{\cos^{2}x}}}$

Now by removing the square root, we get

$\;\;\tt{\rightarrow\;\;L.H.S\:=\:\dfrac{(1\:+\:\sin x)}{\cos x}}$

Now let's divide all the terms of numerator and denominator by cos x , we get

$\;\;\tt{\rightarrow\;\;L.H.S\:=\:\dfrac{\dfrac{1\:+\:\sin x}{\cos x}}{\dfrac{\cos x}{\cos x}}}$

Now this can be written as,

$\;\;\tt{\rightarrow\;\;L.H.S\:=\:\dfrac{\dfrac{1}{\cos x}\:+\:\dfrac{\sin x}{\cos x}}{1}}$

We know that,

• 1/cos x = sec x

• sin x / cos x = tan x

By applying this, we get

$\;\;\tt{\rightarrow\;\;L.H.S\:=\:\dfrac{\sec x\:+\:\tan x}{1}}$

$\;\;\bf{\rightarrow\;\;\orange{L.H.S\:=\:\sec x\:+\:\tan x}}$

Now applying R.H.S. here, we get

$\;\;\bf{\rightarrow\;\;\purple{L.H.S\;=\;R.H.S.\;=\:\sec x\:+\:\tan x}}$

Hence, Proved.

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★ More to know :-

• sin A = cos(90° - A)

• cosec A = sec(90° - A)

• tan A = cot(90° - A)

• cot A = 1/tan A

• cosec A = 1/sin A

• cosec² A = 1 + cot² A

• sin² A + cos² A = 1

• sec² A = 1 + tan² A