Question

tan 9 - tan 27 - tan 63 + tan 81

Answer 1

$$\begin{gathered} Dearstudent, \\ \tan 9°-\tan 27°-\tan 63°+\tan 81° \\ =\tan 9°+\tan 81°-\left(\tan 27°+\tan 63°\right) \\ =\frac{\sin 9°}{\cos 9°}+\frac{\sin 81°}{\cos 81°}-\left(\frac{\sin 27°}{\cos 27°}+\frac{\sin 63°}{\cos 63°}\right) \\ =\frac{\sin 9°\cos 81°+\sin 81°\cos 9°}{\cos 9°.\cos 81°}-\left(\frac{\sin 27°\cos 63°+\sin 63°\cos 27°}{\cos 27°.\cos 63°}\right) \\ =\frac{\sin \left(81°+9°\right)}{\cos 9°.\cos 81°}-\frac{\sin \left(27°+63°\right)}{\cos 27°.\cos 63°} \\ =\frac{\sin 90°}{\cos 9°.\cos 81°}-\frac{\sin 90°}{\cos 27°.\cos 63°} \\ =\frac{1}{\cos 9°.\cos 81°}-\frac{1}{\cos 27°.\cos 63°} \\ =\frac{2}{2\cos 9°.\cos 81°}-\frac{2}{2\cos 27°.\cos 63°} \\ Use2\cos A\cos B=\cos (A+B)+\cos (A-B) \\ =\frac{2}{\left(\cos \left(81°+9°\right)+\cos \left(81°-9°\right)\right)}-\frac{2}{\left(\cos \left(63°+27°\right)+\cos \left(63°-27°\right)\right)} \\ =\frac{2}{\left(\cos 90°+\cos 72°\right)}-\frac{2}{\left(\cos 90°+\cos 36°\right)} \\ =2\left(\frac{1}{\cos 72°}-\frac{1}{\cos 36°}\right) \\ =2\left(\frac{\cos 36°-\cos 72°}{\cos 72°.\cos 72°}\right) \\ =2\times \frac{2\sin 54°\sin 18°}{\cos 72°.\cos 36°} \\ =2\times \frac{2\sin \left(90°-36°\right)\sin \left(90°-18°\right)}{\cos 72°.\cos 36°} \\ =2\times \frac{2\cos 72°.\cos 36°}{\cos 72°.\cos 36°} \\ =2\times 2 \\ =4 \\ Regards \end{gathered}$$