91Ó°ÊÓ

The set of all possible values of \(\alpha\) in \([-\pi, \pi]\) such that \(\sqrt{\frac{1-\sin \alpha}{1+\sin \alpha}}=\sec \alpha-\tan \alpha\) is (a) \([0, \pi / 2)\) (b) \([0, \pi / 2) \cup(\pi / 2, \pi)\) (c) \([-\pi, 0)\) (d) \((-\pi / 2, \pi / 2)\) Solution: (d) Clearly, \(\alpha \neq \pm \pi / 2\) \(\Rightarrow \sec \alpha-\tan \alpha=\frac{1-\sin \alpha}{\cos \alpha}\) and \(\Rightarrow \sqrt{\frac{1-\sin \alpha}{1+\sin \alpha}}=\sqrt{\frac{(1-\sin \alpha)^{2}}{\cos ^{2} \alpha}}=\left|\frac{1-\sin \alpha}{\cos \alpha}\right|=\frac{1-\sin \alpha}{|\cos \alpha|}\) Hence, these will be equal if \(\cos \alpha>0\) i.e., \(-\pi / 2<\) \(\alpha<\pi / 2\)

If \(\tan x / 2=\operatorname{cosec} x-\sin x\), then \(\tan ^{2} x / 2\) is (a) \(2-\sqrt{5}\) (b) \(\sqrt{5}-2\) (c) \((9-4 \sqrt{5})(2+\sqrt{5})\) (d) \((9+4 \sqrt{5})(2-\sqrt{5})\)

Show that \(\sin 12^{\circ} \cdot \sin 48^{\circ} \cdot \sin 54^{\circ}=1 / 8\)

The questions given below consist of an assersion (A) and the reason (R). Use the following key to choose the appropriate answer. (a) If both assertion and reason are correct and reason is the correct explanation of the assertion. (b) If both assertion and reason are correct but reason is not correct explanation of the assertion. (c) If assertion is correct, but reason is incorrect (d) If assertion is incorrect, but reason is correct Now consider the following statements: A. \(\sin 2>\sin 3\) R. If \(x, y \in\left(\frac{\pi}{2}, \pi\right), x\sin y\)

The period of \(\frac{|\sin 4 x|+|\cos 4 x|}{|\sin 4 x-\cos 4 x|+|\sin 4 x+\cos 4 x|}\) is (a) \(\pi / 8\) (b) \(\pi / 2\) (c) \(\pi / 4\) (d) \(\pi\)

If \(\tan \alpha=\sqrt{a}\), where \(a\) is a rational number which is not a perfect square, then which of the following is a rational number? (a) \(\sin 2 \alpha\) (b) \(\tan 2 \alpha\) (c) \(\cos 2 \alpha\) (d) None of these Solution: (c) Given \(\tan \alpha=\sqrt{a}\) $$ \cos 2 \alpha=\frac{1-\tan ^{2} \alpha}{1+\tan ^{2} \alpha}=\frac{1-a}{1+a} $$ \(\Rightarrow\) so it is a rational number $$ \sin 2 \alpha=\frac{2 \tan \alpha}{1+\tan ^{2} \alpha}=\frac{2 \sqrt{a}}{1+a} $$ \(\Rightarrow\) it is an irrational number $$ \tan 2 \alpha=\frac{2 \tan \alpha}{1-\tan ^{2} \alpha}=\frac{2 \sqrt{a}}{1-a} $$ \(\Rightarrow\) it is an irrational number

In a triangle \(P Q R, \angle R=\pi / 2 .\) If \(\tan \left(\frac{P}{2}\right)\) and \(\tan \left(\frac{Q}{2}\right)\) are the roots of the equation \(a x^{2}+b x+c=0,(a \neq 0)\), then (a) \(a+b=c\) (b) \(b+c=a\) (c) \(a+c=b\) (d) \(b=c\)

If \(\theta\) is not an odd multiple of \(\pi / 2\), then prove that \(\tan 9 \theta=\frac{9 \tan \theta-84 \tan ^{3} \theta+126 \tan ^{5} \theta-36 \tan ^{7} \theta+\tan ^{9} \theta}{1-36 \tan ^{2} \theta+126 \tan ^{4} \theta-84 \tan ^{6} \theta+9 \tan ^{8} \theta}\)

If \(a>b>0, y=a \operatorname{cosec} \theta-b \cot \theta\), then for \(0<\theta<\pi\) (a) \(\min y=\sqrt{a^{2}+b^{2}}\) (b) \(\min y=\sqrt{a^{2}-b^{2}}\) (c) y can not become \((a-b)\) for any \(\theta\) (d) y can not become 0 for any \(\theta\)

If \(\alpha\) and \(\beta\) are acute such that \(\tan (\alpha+\beta)\) and \(\tan (\alpha-\beta)\) satisfy the equation \(x^{2}-4 x+1=0\), then \((\alpha, \beta)=\) (a) \(\left(30^{\circ}, 60^{\circ}\right)\) (b) \(\left(45^{\circ}, 45^{\circ}\right)\) (c) \(\left(45^{\circ}, 30^{\circ}\right)\) (d) \(\left(60^{\circ}, 45^{\circ}\right)\)