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.

tan^{-1}x + tan^{-1} y = tan^{-1} [(x + y)/(1 - xy)] , xy < 1

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Answer 1

→ 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 to prove that L.H.S. is equal to R.H.S. There's a simplest method to do it. The method is using assumptions. Firstly we can take the angles given in L.H.S. as some constant values. Then we can convert them in the form of normal Trigonometry. Then we can apply it in R.H.S. to prove the required things.

Let's do it !!

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★ Solution :-

Given to prove,

$\;\;\bf{\mapsto\;\;\orange{\tan^{-1}x\:+\:\tan^{-1}y\;=\;\tan^{-1}\bigg[\dfrac{(x\:+\:y)}{(1\:-\:xy)}\bigg]\:,xy\:<\:1}}$

Let's take some initial values like L.H.S. and R.H.S. from the equation to be proved.

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

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

And also it's given that, xy < 1

» Let tan¯¹ x be equal to a

This will give us,

>> tan¯¹ x = a

>> x = tan a

» Let tan¯¹ y be equal to b

This will give us,

>> tan¯¹ y = b

>> y = tan b

Now let's apply these values in the R.H.S.

We get,

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

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

By identity we know that,

$\;\rm{\leadsto\;\;\tan(\theta\:+\;\phi)\;=\;\dfrac{\tan\theta\:+\:\tan\phi}{1\:-\:\tan\theta\tan\phi}}$

• Here θ = a

• Here ϕ = b

Now by applying this identity in that equation, we get

$\;\;\tt{\rightarrow\;\;R.H.S.\;=\;\tan^{-1}\bigg(\tan(a\:+\:b)\bigg)}$

Now tan¯¹ and tan will cancel out each other giving 1. So we get,

$\;\;\tt{\rightarrow\;\;R.H.S.\;=\;(a\:+\:b)}$

Now on applying the value of a and b, which we assumed earlier, we get

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

Clearly this was what we wanted to proof.

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

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

Hence, proved.

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★ More to know :-

• sin¯¹ 1/x = cosec¯¹ x

• cos¯¹ 1/x = sec¯¹ x

• tan¯¹ 1/x = cot¯¹ x

• sin¯¹ (- x) = - sin¯¹ x

• tan¯¹ (- x) = - tan¯¹ x

• cosec¯¹ (- x) = - cosec¯¹ x