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JEE Main 2022
Trigonometry
Trigonometric Ratio and Identites
Medium

Question

The value of tan9tan27tan63+tan81\tan 9^{\circ}-\tan 27^{\circ}-\tan 63^{\circ}+\tan 81^{\circ} is __________.

Answer: 9

Solution

tan9tan27tan63+tan81=tan9+tan(909)tan27tan(9027)=tan9+cot9tan27cot27=sin9cos9+cos9sin9(sin27cos27+cos27sin27)=sin29+cos29sin9cos9(sin227+cos227cos27sin27)=2sin182sin54=2×4512×45+1=8(5+15+151)=2(2)=4\begin{aligned} & \tan 9^{\circ}-\tan 27^{\circ}-\tan 63^{\circ}+\tan 81^{\circ} \\\\ & =\tan 9^{\circ}+\tan \left(90^{\circ}-9^{\circ}\right)-\tan 27^{\circ}-\tan \left(90^{\circ}-27^{\circ}\right) \\\\ & =\tan 9^{\circ}+\cot 9^{\circ}-\tan 27^{\circ}-\cot 27^{\circ} \\\\ & =\frac{\sin 9^{\circ}}{\cos 9^{\circ}}+\frac{\cos 9^{\circ}}{\sin 9^{\circ}}-\left(\frac{\sin 27^{\circ}}{\cos 27^{\circ}}+\frac{\cos 27^{\circ}}{\sin 27^{\circ}}\right) \\\\ & =\frac{\sin ^2 9^{\circ}+\cos ^2 9^{\circ}}{\sin 9^{\circ} \cos 9^{\circ}}-\left(\frac{\sin ^2 27^{\circ}+\cos ^2 27^{\circ}}{\cos 27^{\circ} \sin 27^{\circ}}\right) \\\\ & =\frac{2}{\sin 18^{\circ}}-\frac{2}{\sin 54^{\circ}} \\\\ & =\frac{2 \times 4}{\sqrt{5}-1}-\frac{2 \times 4}{\sqrt{5}+1} \\\\ & =8\left(\frac{\sqrt{5}+1-\sqrt{5}+1}{5-1}\right)=2(2)=4\end{aligned}

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