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Question: If \(\alpha +\beta =\dfrac{\pi }{2}\) and \(\beta +\gamma =\alpha \), then \(\tan \alpha \) equals: ...

If α+β=π2\alpha +\beta =\dfrac{\pi }{2} and β+γ=α\beta +\gamma =\alpha , then tan⁡α\tan \alpha equals:
(a) 2(tan⁡β+tan⁡γ)2\left( \tan \beta +\tan \gamma \right)
(b) tan⁡β+tan⁡γ\tan \beta +\tan \gamma
(c) tan⁡β+2tan⁡γ\tan \beta +2\tan \gamma
(d) 2tan⁡β+tan⁡γ2\tan \beta +\tan \gamma

Explanation

Solution

We have given the equation α+β=π2\alpha +\beta =\dfrac{\pi }{2}, subtracting β\beta on both the sides and then take tan on both the sides will give the equation tan⁡α=tan⁡(π2−β)\tan \alpha =\tan \left( \dfrac{\pi }{2}-\beta \right) further solving this equation will give you a relation between tan⁡α&tan⁡β\tan \alpha \And \tan \beta . Mark it as eq. (1) then solve the other equation i.e. β+γ=α\beta +\gamma =\alpha by subtracting β\beta on both the sides after that take tan on both the sides of the equation then simplify it further and use the eq. (1) to simplify it further and will get you a relation between tan⁡α,tan⁡β&tan⁡γ\tan \alpha ,\tan \beta \And \tan \gamma

Complete step by step answer:
We have given the following equation,
α+β=π2\alpha +\beta =\dfrac{\pi }{2}
Subtracting β\beta on both the sides we get,
α=π2−β\alpha =\dfrac{\pi }{2}-\beta
Taking tan on both the sides of the above equation we get,
tan⁡α=tan⁡(π2−β)\tan \alpha =\tan \left( \dfrac{\pi }{2}-\beta \right)
We know from the trigonometric ratios that tan⁡(π2−β)=cot⁡β\tan \left( \dfrac{\pi }{2}-\beta \right)=\cot \beta using this relation in the above expression we get,
tan⁡α=cot⁡β\tan \alpha =\cot \beta
Using the trigonometric relation of cot⁡β=1tan⁡β\cot \beta =\dfrac{1}{\tan \beta } in the above equation we get,
tan⁡α=1tan⁡β\tan \alpha =\dfrac{1}{\tan \beta }
Cross multiplying the above equation we get,
tan⁡α(tan⁡β)=1\tan \alpha \left( \tan \beta \right)=1……….. Eq. (1)
We have also given another equation as follows:
β+γ=α\beta +\gamma =\alpha
Subtracting β\beta on both the sides of the above equation we get,
γ=α−β\gamma =\alpha -\beta
Now, taking tan on both the sides of the above equation we get,
tan⁡γ=tan⁡(α−β)\tan \gamma =\tan \left( \alpha -\beta \right)
We know the identity of tan⁡(α−β)=tan⁡α−tan⁡β1+tan⁡αtan⁡β\tan \left( \alpha -\beta \right)=\dfrac{\tan \alpha -\tan \beta }{1+\tan \alpha \tan \beta } using this identity in the above equation we get,
tan⁡γ=tan⁡α−tan⁡β1+tan⁡αtan⁡β\tan \gamma =\dfrac{\tan \alpha -\tan \beta }{1+\tan \alpha \tan \beta }
Using eq. (1) we can substitute the value of tan⁡αtan⁡β=1\tan \alpha \tan \beta =1 in the above equation we get,
tan⁡γ=tan⁡α−tan⁡β1+1 ⇒tan⁡γ=tan⁡α−tan⁡β2 \begin{aligned} & \tan \gamma =\dfrac{\tan \alpha -\tan \beta }{1+1} \\\ & \Rightarrow \tan \gamma =\dfrac{\tan \alpha -\tan \beta }{2} \\\ \end{aligned}
Cross multiplying the above equation we get,
2tan⁡γ=tan⁡α−tan⁡β2\tan \gamma =\tan \alpha -\tan \beta
Adding tan⁡β\tan \beta on both the sides of the above equation we get,
2tan⁡γ+tan⁡β=tan⁡α2\tan \gamma +\tan \beta =\tan \alpha
Now, compare the relation that we are getting with the options given in the question.

So, the correct answer is “Option c”.

Note: You might have thought that in simplifying the equation β+γ=α\beta +\gamma =\alpha , why we have subtracted β\beta on both the sides and then take tan on both the sides instead we can subtract γ\gamma and then take tan on both the sides also.
The answer is let us subtract γ\gamma on both the sides of the equation β+γ=α\beta +\gamma =\alpha we get,
β=α−γ\beta =\alpha -\gamma
Taking tan on both the sides of the above equation we get,
tan⁡β=tan⁡(α−γ) ⇒tan⁡β=tan⁡α−tan⁡γ1+tan⁡αtan⁡γ \begin{aligned} & \tan \beta =\tan \left( \alpha -\gamma \right) \\\ & \Rightarrow \tan \beta =\dfrac{\tan \alpha -\tan \gamma }{1+\tan \alpha \tan \gamma } \\\ \end{aligned}
Now, you can see that we cannot use the relation in eq. (1) i.e. tan⁡α(tan⁡β)=1\tan \alpha \left( \tan \beta \right)=1 this is the reason why we haven’t simplified the equation in this way.