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.

3tan^{-1}x = sin^{-1} [(2x)/(1 + x²)] , |x| ≤ 1

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Answer 1

★ Correct Qúestion :-

Prove the following identity :-

$\;\bf{\odot\;\;\green{2\tan^{-1}\:x\;=\;\sin^{-1}\dfrac{2x}{1\:+\:x^{2}}}}$

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★ Concept :-

Here the concept of Trigonometric Identities have been used. This is the question of Inverse Trigonometry. So the easiest method to solve such questions is using an assumption. Just assume a variable with a key value. Then we can initialise the equation and thus find the answer. Also we shall be using different trignometric equations in this.

Let's do it !!

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★ Solution :-

Let tan¯¹ x = y

This will give us, x = tan y

Now let's apply this value of x in the R.H.S of the equation to be proved .

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

By applying value here, we get

$\;\tt{\rightarrow\;\;R.H.S.\;=\;\sin^{-1}\dfrac{2(\tan y)}{1\:+\:(\tan y)^{2}}}$

$\;\tt{\rightarrow\;\;R.H.S.\;=\;\sin^{-1}\dfrac{2\tan y}{1\:+\:\tan^{2}y}}$

Now by another trignometric identity, we know that,

$\;\sf{\leadsto\;\;\sin 2x\;=\;\dfrac{2\tan x}{1\:+\:\tan^{2} x}}$

• Here x = y

By applying this, we get

$\;\tt{\rightarrow\;\;R.H.S.\;=\;\sin^{-1}(\sin 2y)}$

Now sin¯¹ and sin will cancel out each other.

$\;\tt{\rightarrow\;\;R.H.S.\;=\;1\:\times\:2y}$

$\;\bf{\rightarrow\;\;R.H.S.\;=\;2y}$

Now let's apply the value of y. We get,

$\;\bf{\rightarrow\;\;R.H.S.\;=\;2(\tan^{-1} x)}$

$\;\bf{\rightarrow\;\;\blue{R.H.S.\;=\;2\tan^{-1} x}}$

Let's now compare LHS and RHS.

Clearly, LHS = RHS.

$\;\bf{\rightarrow\;\;\red{L.H.S.\;=\;R.H.S.\;=\;2\tan^{-1}x}}$

By applying initial values, this will give us,

$\;\bf{\mapsto\;\;\purple{2\tan^{-1}\:x\;=\;\sin^{-1}\dfrac{2x}{1\:+\:x^{2}}}}$

Hence, Proved.

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★ More to know :-

• sin 2x = 2 sin x cos x

• cos 2x = 2 cos² x - 1 or 1 - 2 sin² x

• sin 3x = 3 sin x - 4 sin³ x

• cos 3x ° 4 cos³ x - 3 cos x

• sin² x - sin² y = sin(x + y) • sin(x - y)

• 2 cos C cos D = cos(C + D) + cos(C - D)

• -2 sin C sin D = cos(C + D) - cos(C - D)

• 2 cos C sin D = sin(C + D) - sin(C - D)

• 2 sin C cos D = sin(C + D) + sin(C - D)

Answer 2

Answer:

Answer in the attachment.