Trigonometric Ratio and Identities MCQ Questions & Answers in Trigonometry | Maths

Clearly, $$\alpha \ne \pm \frac{\pi }{2}.$$
$$\eqalign{ & \sec\alpha - \tan\alpha = \frac{{1 - \sin \alpha }}{{\cos \alpha }}\,\,{\text{and}} \cr & \sqrt {\frac{{1 - \sin \alpha }}{{1 + \sin \alpha }}} = \sqrt {\frac{{{{\left( {1 - \sin \alpha } \right)}^2}}}{{{{\cos }^2}\alpha }}} = \left| {\frac{{1 - \sin \alpha }}{{\cos \alpha }}} \right| = \frac{{1 - \sin \alpha }}{{\left| {\cos \alpha } \right|}}. \cr} $$
Hence, these will be equal if $$\cos\alpha > 0,\,{\text{i}}{\text{.e}}{\text{., }} - \frac{\pi }{2} < \alpha < \frac{\pi }{2}.$$

$$\eqalign{ & \frac{{\sin x - \sin 3x}}{{{{\sin }^2}x - {{\cos }^2}x}} = \frac{{\sin 3x - \sin x}}{{{{\cos }^2}x - {{\sin }^2}x}} \cr & = \frac{{2\cos \frac{{3x + x}}{2} \cdot \sin \frac{{3x - x}}{2}}}{{\cos 2x}} = \frac{{2\cos 2x \cdot \sin x}}{{\cos 2x}} \cr & = 2\sin x \cr} $$

Let $${f_k}\left( x \right) = \frac{1}{k}\left( {{{\sin }^k}x + {{\cos }^k}x} \right)$$
Consider
$$\eqalign{ & {f_4}\left( x \right) - {f_6}\left( x \right) = \frac{1}{4}\left( {{{\sin }^4}x + {{\cos }^4}x} \right) - \frac{1}{6}\left( {{{\sin }^6}x + {{\cos }^6}x} \right) \cr & = \frac{1}{4}\left[ {1 - 2{{\sin }^2}x{{\cos }^2}x} \right] - \frac{1}{6}\left[ {1 - 3{{\sin }^2}x{{\cos }^2}x} \right] \cr & = \frac{1}{4} - \frac{1}{6} \cr & = \frac{1}{{12}} \cr} $$

$$\sin \theta = \frac{{2 \pm \sqrt {4 - 4\lambda } }}{2} = 1 \pm \sqrt {1 - \lambda } .$$       For real values, $$1 - \lambda \geqslant 0,\,\,{\text{i}}{\text{.e}}{\text{., }}\lambda \leqslant 1.$$
As $$ - 1 \leqslant \sin\theta \leqslant 1,\sin\theta = 1 - \sqrt {1 - \lambda } .$$
From the question, $$\sin\theta > - \frac{1}{2}.\,{\text{Thus }} - \frac{1}{2} < 1 - \sqrt {1 - \lambda } \leqslant 1$$
$$\eqalign{ & {\text{or, }} - \frac{3}{2} < - \sqrt {1 - \lambda } \leqslant 0 \cr & \Rightarrow \,\,\sqrt {1 - \lambda } < \frac{3}{2} \cr & \Rightarrow \,\,1 - \lambda < \frac{9}{4} \cr & \Rightarrow \,\,\lambda > - \frac{5}{4}. \cr} $$

Numerator of given expression
$$\eqalign{ & = \left( {\sin 8x + \sin 6x} \right) + 6\left( {\sin 6x + \sin 4x} \right) + 12\left( {\sin 4x + \sin 2x} \right) \cr & = 2\sin 7x\cos x + 12\sin 5x\cos x + 24\sin 3x\cos x \cr & = 2\cos x\left( {\sin 7x + 6\sin 5x + 12\sin 3x} \right) \cr} $$
Hence, given expression
$$ = \frac{{2\cos x\left( {\sin 7x + 6\sin 5x + 12\sin 3x} \right)}}{{\left( {\sin 7x + 6\sin 5x + 12\sin 3x} \right)}} = 2\cos x$$

$$\eqalign{ & \Rightarrow {\text{Period of}}\,\,\sin x = 2\pi \cr & \Rightarrow {\text{Period of}}\,\,{\sin ^3}x = 2\pi \cr & {\text{Period of}}\,\,\left| {{{\sin }^3}x} \right| = \pi \cr & \Rightarrow {\text{Period of}}\,\,\left| {{{\sin }^3}\frac{x}{2}} \right| = 2\pi \cr & {\text{Period of}}\,\,{\cos ^5}x = 2\pi \cr & \Rightarrow {\text{Period of}}\,\,\left| {{{\cos }^5}x} \right| = \pi \cr & \Rightarrow {\text{Period of}}\,\,\left| {{{\cos }^5}\frac{x}{5}} \right| = 5\pi \cr} $$
Thus required period $$ = {\text{LCM of }}2\pi \,\,\& \,\,5\pi = 10\pi $$

$$\eqalign{ & {\left( {\cos {9^ \circ } - \sin{9^ \circ }} \right)^2} = 1 - \sin {18^ \circ } \cr & {\left( {\cos {9^ \circ } - \sin{9^ \circ }} \right)^2} = 1 - \frac{{\sqrt 5 - 1}}{4} = \frac{{5 - \sqrt 5 }}{4};\,{\text{also }}\cos {9^ \circ } > \sin{9^ \circ }. \cr} $$

$$\eqalign{ & {\text{L}}{\text{.H}}{\text{.S}} = \tan \theta + \frac{{\tan \theta + \sqrt 3 }}{{1 - \sqrt 3 \tan \theta }} + \frac{{\tan \theta - \sqrt 3 }}{{1 + \sqrt 3 \tan \theta }} \cr & {\text{L}}{\text{.H}}{\text{.S}} = \frac{{\left\{ {\tan \theta - 3{{\tan }^3}\theta + \left( {\tan \theta + \sqrt 3 } \right)\left( {1 + \sqrt 3 \tan \theta } \right) + \left( {\tan \theta - \sqrt 3 } \right)\left( {1 - \sqrt 3 \tan \theta } \right)} \right\}}}{{1 - 3{{\tan }^2}\theta }} \cr & {\text{L}}{\text{.H}}{\text{.S}} = \frac{{3\left( {3\tan \theta - {{\tan }^3}\theta } \right)}}{{1 - 3{{\tan }^2}\theta }} = 3\tan 3\theta ; \cr & \therefore \,\,\,\,k = 3. \cr} $$

$$\eqalign{ & D \geqslant 0 \cr & \Rightarrow \,\,{\cos ^2}p - 4\sin p \cdot \left( {\cos p - 1} \right) \geqslant 0 \cr & {\text{or, }}{\cos ^2}p + 4\sin p\left( {1 - \cos p} \right) \geqslant 0\,\,\,\,\,\,\,.....\left( 1 \right) \cr} $$
$${\text{As }}{\cos ^2}p \geqslant 0\,\,{\text{and }}1 - \cos p \geqslant 0,\left( 1 \right)$$       will hold for all $$\sin p \geqslant 0.\,{\text{So, }}p \in \left[ {0,\pi } \right].$$

$$\eqalign{ & {\tan ^2}\theta \,{\sec ^2}\theta \left( {{{\cot }^2}\theta - {{\cos }^2}\theta } \right) \cr & = {\sec ^2}\theta \left( {{{\tan }^2}\theta \,{{\cot }^2}\theta - {{\tan }^2}\theta \,{{\cos }^2}\theta } \right) \cr & = {\sec ^2}\theta \left( {1 - \frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}{{\cos }^2}\theta } \right) = {\sec ^2}\theta \left( {1 - {{\sin }^2}\theta } \right) \cr & = {\sec ^2}\theta \cdot {\cos ^2}\theta = 1 \cr} $$