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.

2 tan^{-1} x = tab^{-1} [2x/(1 - x²)] , -1 < x < 1

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Answer 1

Answer:

→ Let α=tan

−1

x,β=tan

−1

y

As tan(α+β)=

1−tanαtanβ

tanα+tanβ

⇒tan(α+β)=

1−xy

x+y

⇒α+β=tan

−1

(

1−xy

x+y

)= tan

−1

x+tan

−1

y

Answer 2

★ Concept :-

Here the concept of Inverse Trigonometry has been used. We see that we are given an equation and we have to prove that L.H.S. is equal to R.H.S. This can be done by simply assuming the main terms. From that we can get a value which we can apply in the R.H.S. and thus find the answer.

Let's do it !!

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★ Solution :-

Given to prove,

$\;\;\bf{\mapsto\;\;\green{2\tan^{-1}x\;=\;\tan^{-1}\bigg[\dfrac{2x}{(1\:-\:x^{2})}\bigg],\;-1\:<\:x\:<\:1}}$

This is the appropriate qúestion.

From this we get,

$\;\;\sf{\odot\;\;\blue{L.H.S.\;=\;2\tan^{-1}x}}$

$\;\;\sf{\odot\;\;\orange{R.H.S.\;=\;tan^{-1}\bigg[\dfrac{2x}{(1\:-\:x^{2})}\bigg]}}$

And, -1 < x < 1

• Let tan¯¹ x be equal to y.

Then we get,

>> tan¯¹ x = y

>> x = tan y

Now let's apply the value of x in the R.H.S., then we get,

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

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

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

By identity, we know that :

$\;\rm{\leadsto\;\;\tan 2\theta\;=\;\dfrac{2\tan \theta}{1\:-\:\tan^{2}\theta}}$

• Here θ = y

By applying the values, we get

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

Here tan¯¹ and tan will cancel each other giving 1. So, by using this we get

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

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

Now by applying the value of y, we get

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

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

Thus we get the value of L.H.S. and R.H.S. to be equal.

>> L.H.S. = R.H.S.

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

• Hence, proved

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★ More to know :-

• sin 2A = 2 sin A cos A

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

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

• sin² A + cos² A = 1

• sec² A = 1 + tan² A

• cosec² A = 1 + cot² A