Question 18

Find $$\tan^4 \alpha + \tan^4 \gamma$$ using the information given below:
$$\tan(\theta - \gamma) = \frac{1}{\sqrt{2}}, \tan \theta \tan \gamma = \tan^2 \alpha$$

Solution

Given that,

$$\tan\left(\theta\ -\gamma\ \right)=\frac{1}{\sqrt{\ 2}}$$

$$\frac{\tan\theta-\tan\gamma}{1+\tan\theta\ \tan\gamma\ }\ =\frac{1}{\sqrt{\ 2}}$$

$$\frac{\tan\theta-\tan\gamma}{1+\tan^2\alpha\ }\ =\frac{1}{\sqrt{\ 2}}$$

$$\frac{\tan\theta-\tan\gamma}{\sec^2\alpha\ }\ =\frac{1}{\sqrt{\ 2}}$$

$$\ \tan\theta\ -\tan\gamma\ =\frac{1}{\sqrt{\ 2}\cos^2\alpha\ }$$

$$\ \frac{\tan^2\alpha\ }{\tan\gamma\ }-\tan\gamma\ =\frac{1}{\sqrt{\ 2}\cos^2\alpha\ }$$

$$\ \tan^2\alpha\ -\tan^2\gamma\ =\frac{\tan\gamma\ }{\sqrt{\ 2}\cos^2\alpha\ }$$

Squaring on both sides,

$$\tan^4\alpha\ +\tan^4\gamma-2\ \tan^2\alpha\ \tan^2\gamma\ =\frac{\tan^2\gamma\ }{2\cos^4\alpha\ }$$

$$\tan^4\alpha\ +\tan^4\gamma\ =2\ \tan^2\alpha\ \tan^2\gamma+\frac{\tan^2\gamma\ }{2\cos^4\alpha\ }$$

$$\tan^4\alpha\ +\tan^4\gamma\ =\tan^2\gamma\ \left(2\ \tan^2\alpha\ +\frac{1}{2\cos^4\alpha\ }\right)$$

$$\tan^4\alpha\ +\tan^4\gamma\ =\tan^2\gamma\ \left(\ \frac{4\sin^2\alpha\ \cos^2\alpha}{2\cos^4\alpha}\ +\frac{1}{2\cos^4\alpha\ }\right)$$

$$\tan^4\alpha\ +\tan^4\gamma\ =\frac{\tan^2\gamma}{2}\ \left(\ \frac{4\sin^2\alpha\ \cos^2\alpha+1}{\cos^4\alpha}\ \right)$$

$$=\frac{\tan^2\gamma}{2}\ \left(\ \frac{6\sin^2\alpha\ \cos^2\alpha+1-2\sin^2\alpha\ \cos^2\alpha}{\cos^4\alpha}\ \right)$$

$$=\frac{\tan^2\gamma}{2}\ \left(\ \frac{6\sin^2\alpha\ \cos^2\alpha+\left(\sin^2\alpha\ +\cos^2\alpha\right)^2-2\sin^2\alpha\ \cos^2\alpha}{\cos^4\alpha}\ \right)$$

$$=\frac{\tan^2\gamma}{2}\ \left(\ \frac{6\sin^2\alpha\ \cos^2\alpha+\sin^4\alpha\ +\cos^4\alpha+2\sin^2\alpha\ \cos^2\alpha-2\sin^2\alpha\ \cos^2\alpha}{\cos^4\alpha}\ \right)$$

$$=\frac{\tan^2\gamma}{2}\ \left(\ \frac{6\sin^2\alpha\ \cos^2\alpha+\sin^4\alpha\ +\cos^4\alpha}{\cos^4\alpha}\ \right)$$

$$=\frac{\tan^2\gamma}{2}\ \left(\ 6\tan^2\alpha+\tan^4\alpha\ +1\ \right)$$

$$=\frac{\tan^2\gamma}{2}\ \left(\ \left(\tan^2\alpha+3\right)^2\ -8\ \right)$$

Option (A) is the answer.


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