Question

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

Proof the given identity.

[tan/(1 - cotx)] + [cot/(1 - tanx)] = 1 + secxcosecx

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Answer 1

★ Concept :-

Here the concept of Trigonometric Identities has been used. We see that we are given an equation where we need to prove that if L.H.S. is equal to R.H.S. So firstly here we shall simplify L.H.S. and then try to bring L.H.S. in the form of R.H.S. In doing so, we shall use different algebraic and trignometric identities. Then we will apply formula and thus prove the given equation.

Let's do it !!

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★ Solution :-

Given to prove,

$\;\bf{\mapsto\;\;\red{\bigg[\dfrac{\tan x}{(1\:-\:\cot x)}\bigg]\;+\;\bigg[\dfrac{\cot x}{(1\:-\:\tan x)}\bigg]\;=\;1\:+\:\sec x\cosec x}}$

This is the appropriate qúestion.

From this we get,

$\;\sf{\leadsto\;\;\green{L.H.S.\;=\;\bigg[\dfrac{\tan x}{(1\:-\:\cot x)}\bigg]\;+\;\bigg[\dfrac{\cot x}{(1\:-\:\tan x)}\bigg]}}$

$\;\sf{\leadsto\;\;\blue{R.H.S.\;=\;1\:+\:\sec x\cosec x}}$

Now let's begin simplification of L.H.S.

So,

$\;\tt{\rightarrow\;\;L.H.S.\;=\;\bigg[\dfrac{\tan x}{(1\:-\:\cot x)}\bigg]\;+\;\bigg[\dfrac{\cot x}{(1\:-\:\tan x)}\bigg]}$

We know that,

• tan A = (sin A)/(cos A)

• cot A = (cos A)/(sin A)

Here A = x

By applying this, we get

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

$\;\tt{\rightarrow\;\;L.H.S.\;=\;\bigg[\dfrac{\dfrac{\sin x}{\cos x}}{\dfrac{\sin x\:-\:\cos x}{\sin x}}\bigg]\;+\;\bigg[\dfrac{\dfrac{\cos x}{\sin x}}{\dfrac{\cos x\:-\:\sin x}{\cos x}}\bigg]}$

Now by changing sign, we get

$\;\tt{\rightarrow\;\;L.H.S.\;=\;\bigg[\dfrac{\dfrac{\sin x}{\cos x}}{\dfrac{\sin x\:-\:\cos x}{\sin x}}\bigg]\;-\;\bigg[\dfrac{\dfrac{\cos x}{\sin x}}{\dfrac{\sin x\:-\:\cos x}{\cos x}}\bigg]}$

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

Now let's take the terms common.

$\;\tt{\rightarrow\;\;L.H.S.\;=\;\dfrac{1}{(\sin x\:-\:\cos x)}\bigg[\dfrac{\sin^{2}x}{\cos x}\:-\:\dfrac{\cos^{2}x}{\sin x}\bigg]}$

Now taking the L.C.M. of the fraction, we get

$\;\tt{\rightarrow\;\;L.H.S.\;=\;\dfrac{1}{(\sin x\:-\:\cos x)}\bigg[\dfrac{\sin^{3}x\:-\:\cos^{3}x}{\cos x\sin x}\bigg]}$

We know that,

• a³ - b³ = (a - b)(a² + b² + 2ab)

Here a = sin³ x and b = cos³ x

By applying this identity, we get

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

We know that,

• sin² A + cos² A = 0

Here A = x

By applying this, we get

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

Now on opening the bracket, we get

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

On cancelling the like terms from numerator and denominator, we get

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

Now this fraction can be written as,

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

We know that,

• sec A = 1/(cos A)

• cosec A = 1/(sin A)

By applying these here, we get

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

$\;\bf{\rightarrow\;\;\orange{L.H.S.\;=\;1\;+\;\sec x\:cosec\:x}}$

Clearly, L.H.S. = R.H.S.

So,

$\;\bf{\rightarrow\;\;\purple{R.H.S.\;=\;L.H.S.\;=\;1\;+\;\sec x\:cosec\:x}}$

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★ More to know :-

• sec² A = 1 + tan² A

• cosec A = 1 + cot² A

• sin A = cos(90° - A)

• cosec A = sec(90° - A)

• tan A = cot(90° - A)