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Question Number 68397 by Faradtimmy last updated on 10/Sep/19

Answered by $@ty@m123 last updated on 10/Sep/19

$$\left(a\right)LHS=\sqrt{3}\cos \theta -\sin \theta$$$$=2(\frac{\sqrt{3}}{2}\cos \theta -\frac{1}{2}\sin \theta )$$$$=2(\cos 30\cos \theta -\sin 30\sin \theta )$$$$=2\cos (30+\theta )$$$$Pl.checkthequestion.$$$$\left(b\right)\sqrt{\frac{1-\sin \theta }{1+\sin \theta }}$$$$=\sqrt{\frac{1-\sin \theta }{1+\sin \theta }\times \frac{1-\sin \theta }{1-\sin \theta }}$$$$=\sqrt{\frac{{(1-\sin \theta )}^2}{1-{\sin }^2\theta }}$$$$=\sqrt{{\left(\frac{1-\sin \theta }{\cos \theta }\right)}^2}$$$$=\sqrt{{(\sec \theta -\tan \theta )}^2}$$$$=-(\sec \theta -\tan \theta ),\because {90}^{\mathrm{o}}<\theta ⩽{180}^{\mathrm{o}}$$$$Similarly,$$$$\sqrt{\frac{1+\sin \theta }{1-\sin \theta }}=-(\sec \theta +\tan \theta )$$$$\therefore thegivenexpression$$$$=-(\sec \theta -\tan \theta )-(\sec \theta +\tan \theta )$$$$=-2\sec \theta$$

Answered by $@ty@m123 last updated on 10/Sep/19

$$\left(c\right)\frac{\sec 8\theta -1}{\sec 4\theta -1}$$$$=\frac{\sec 8\theta (1-\cos 8\theta )}{\sec 4\theta (1-\cos 4\theta )}$$$$=\frac{\cos 4\theta .2\sin {}^{2}4\theta }{\cos 8\theta .2\sin {}^{2}2\theta }$$$$=\frac{2\sin 4\theta \cos 4\theta \sin 4\theta }{2\cos 8\theta \sin {}^{2}2\theta }$$$$=\frac{\sin 8\theta .2\sin 2\theta \cos 2\theta }{2\cos 8\theta \sin {}^{2}2\theta }$$$$=\frac{\tan 8\theta }{\tan 2\theta }$$$$2.\left(a\right)\tan 9-\tan 27-\tan 63+\tan 81$$$$=\tan 9-\tan 27-\cot 27+\cot 9$$$$=\frac{\sin 9}{\cos 9}+\frac{\cos 9}{\sin 9}-(\frac{\sin 27}{\cos 27}+\frac{\cos 27}{\sin 27})$$$$=\frac{1}{\cos 9\sin 9}-\frac{1}{\cos 27\sin 27}$$$$=\frac{2}{\sin 18}-\frac{2}{\sin 54}$$$$=2\left(\frac{\sin 54-\sin 18}{\sin 18\sin 54}\right)$$$$=\frac{4\cos 36\sin 18}{\sin 18\sin 54}$$$$=\frac{4\cos 36}{\sin 54}$$$$=\frac{4\cos 36}{\cos 36}$$$$=4$$

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