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Question: If \({{\tan }^{-1}}2+{{\tan }^{-1}}3+\theta =\pi \) , find the value of \(\theta \) ?...

If tan⁡−12+tan⁡−13+θ=π{{\tan }^{-1}}2+{{\tan }^{-1}}3+\theta =\pi , find the value of θ\theta ?

Explanation

Solution

In the given question, we need to find the value of one variable or as per the question we can say an angle theta. We are given an equation involving inverse trigonometric function tangent at some angles and adding theta to it makes its value equivalent to pi.

Complete step-by-step answer:
Now, in order to find the variable θ\theta what we will do is we will shift θ\theta to right-hand side of the equation we get tan⁡−12+tan⁡−13=π−θ{{\tan }^{-1}}2+{{\tan }^{-1}}3=\pi -\theta and then we can apply the inverse trigonometric identity involving inverse of tangent function with angles A and B as 2 and 3 respectively.
So, now we know that tan⁡−1(A)+tan⁡−1(B)=tan⁡−1(A+B1−AB){{\tan }^{-1}}\left( A \right)+{{\tan }^{-1}}\left( B \right)={{\tan }^{-1}}\left( \dfrac{A+B}{1-AB} \right)
Now, putting the values of A and B we get,
tan⁡−1(2)+tan⁡−1(3)=tan⁡−1(2+31−2×3){{\tan }^{-1}}\left( 2 \right)+{{\tan }^{-1}}\left( 3 \right)={{\tan }^{-1}}\left( \dfrac{2+3}{1-2\times 3} \right)
So, now according to our equation we get,
tan⁡−1(2+31−2×3)=π−θ ⇒tan⁡−1(5−5)=π−θ ⇒tan⁡−1(−1)=π−θ ⇒−1=tan⁡(π−θ) \begin{aligned} & {{\tan }^{-1}}\left( \dfrac{2+3}{1-2\times 3} \right)=\pi -\theta \\\ & \Rightarrow {{\tan }^{-1}}\left( \dfrac{5}{-5} \right)=\pi -\theta \\\ & \Rightarrow {{\tan }^{-1}}\left( -1 \right)=\pi -\theta \\\ & \Rightarrow -1=\tan \left( \pi -\theta \right) \\\ \end{aligned}
Now again, we know that tan⁡(π−θ)=−tan⁡θ\tan \left( \pi -\theta \right)=-\tan \theta . We get this identity when we check the angles at various trigonometric functions and divide the six functions in four quadrants according to various angles. This tan⁡(π−θ)=−tan⁡θ\tan \left( \pi -\theta \right)=-\tan \theta identity holds as the angle involved with tangent lies in the second quadrant and tangent function is negative in that quadrant.
Therefore,
−1=−tan⁡θ ⇒θ=π4 \begin{aligned} & -1=-\tan \theta \\\ & \Rightarrow \theta =\dfrac{\pi }{4} \\\ \end{aligned}
Therefore, the value of angle theta as per the given equation is θ=π4\theta =\dfrac{\pi }{4} .

Note: In the given question we need to be careful while applying the identities as they are similar to other functions identities and also inverse trigonometric identities are similar to trigonometric function identities. Apart from this we must know in which quadrant which function is having which sign as it plays a very important role in attaining the right answer.