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

Given that in $$\Delta PQR,\angle R = \frac{\pi }{2}$$
$$\eqalign{ & \Rightarrow \,\,\angle P + \angle Q = \frac{\pi }{2} \cr & \Rightarrow \,\,\frac{{\angle P}}{2} + \frac{{\angle Q}}{2} = \frac{\pi }{4} \cr} $$
Also $$\tan \frac{P}{2}{\text{ and tan}}\frac{Q}{2}$$   are roots of the equation
$$\eqalign{ & a{x^2} + bx + c = 0\left( {a \ne 0} \right) \cr & \therefore \,\,\tan \frac{P}{2} + \tan \frac{Q}{2} = - \frac{b}{a};\tan \frac{P}{2}\tan \frac{Q}{2} = \frac{c}{a} \cr} $$
Now consider, $$\tan \left( {\frac{{P + Q}}{2}} \right) = \frac{{\tan \frac{P}{2} + \tan \frac{Q}{2}}}{{1 - \tan \frac{P}{2}\tan \frac{Q}{2}}}$$
$$\eqalign{ & \Rightarrow \,\,\tan \frac{\pi }{4} = \frac{{ - \frac{b}{a}}}{{1 - \frac{c}{a}}} \cr & \Rightarrow \,\,1 - \frac{c}{a} = - \frac{b}{a} \cr & \Rightarrow \,\,a - c = - b \cr & \Rightarrow \,\,a + b = c \cr} $$

$$\eqalign{ & {\text{In a}}\,\Delta ABC,{\text{we have}} \cr & \sin A - \cos B = \cos C \cr & \Rightarrow \sin A = \cos B + \cos C \cr & \Rightarrow 2\sin \frac{A}{2} \cdot \cos \frac{A}{2} \cr & = 2\cos \left( {\frac{{B + C}}{2}} \right) \cdot \cos \left( {\frac{{B - C}}{2}} \right)\,\,\,\left[ {\because \sin 2A = 2\sin A \cdot \cos A} \right] \cr & {\text{and}}\,\,\cos B + \cos C = 2\cos \left( {\frac{{B + C}}{2}} \right) \cdot \cos \left( {\frac{{B - C}}{2}} \right) \cr & \Rightarrow 2\sin \frac{A}{2} \cdot \cos \frac{A}{2} = 2\cos \left( {{{90}^ \circ } - \frac{A}{2}} \right) \cdot \cos \left( {\frac{{B - C}}{2}} \right) \cr & \left[ {\because A + B + C = {{180}^ \circ } \Rightarrow \left( {\frac{{B + C}}{2}} \right) = {{90}^ \circ } - \frac{A}{2}} \right] \cr & \Rightarrow 2\sin \frac{A}{2} \cdot \cos \frac{A}{2} = 2\sin \frac{A}{2} \cdot \cos \left( {\frac{{B - C}}{2}} \right)\,\,\left[ {\because \cos \left( {{{90}^ \circ } - \theta } \right) = \sin \theta } \right] \cr & \Rightarrow \cos \frac{A}{2} = \cos \left( {\frac{{B - C}}{2}} \right) \cr & \Rightarrow \frac{A}{2} = \frac{{B - C}}{2} \cr & \Rightarrow A + C = B\,\,\,.....\left( {\text{i}} \right) \cr & {\text{Also}},A + C = {180^ \circ } - B\,\,\,.....\left( {{\text{ii}}} \right) \cr & {\text{So}},{180^ \circ } - B = B \cr & \Rightarrow 2B = {180^ \circ }\,\,\,\therefore B = {90^ \circ } \cr} $$

In a convex quadrilateral no angle is greater than $${180^ \circ }$$
$$\eqalign{ & {\text{Here, }}\tan A = \frac{5}{{12}}.\,{\text{So, }}0 < A < \frac{\pi }{2}\,{\text{and }}\frac{\pi }{2} < C < \pi \,\,\left( {\because \,\,A + C = {{180}^ \circ }} \right) \cr & \therefore \,\,\tan \left( {\pi - C} \right) = \frac{5}{{12}},\,{\text{i}}{\text{.e}}{\text{., }}\tan C = - \frac{5}{{12}}\,\,\,\,\,\,\therefore \,\,\cos C = - \frac{{12}}{{13}}. \cr & {\text{Also, }}\cos B = - \frac{3}{5}.\,{\text{So, }}\frac{\pi }{2} < B < \pi \,\,{\text{and }}0 < D < \frac{\pi }{2}\,\,\,\,\,\,\left( {\because \,\,B + D = {{180}^ \circ }} \right) \cr & \therefore \,\,\cos \left( {\pi - D} \right) = - \frac{3}{5},\,{\text{i}}{\text{.e}}{\text{., }}\cos D = \frac{3}{5}\,\,\,\,\,\,\therefore \,\,\tan D = \frac{4}{3}. \cr} $$
∴ the required equation is $${x^2} - \left( { - \frac{{12}}{{13}} + \frac{4}{3}} \right)x + \left( { - \frac{{12}}{{13}}} \right) \cdot \frac{4}{3} = 0.$$

The expression $$ = \frac{{\left| {\cos A + \sin A} \right| + \left| {\cos A - \sin A} \right|}}{{\left| {\cos A + \sin A} \right| - \left| {\cos A - \sin A} \right|}}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{\cos A + \sin A + \cos A - \sin A}}{{\cos A + \sin A - \left( {\cos A - \sin A} \right)}}$$         because $$ - \frac{\pi }{4} < A < \frac{\pi }{4}$$   and in this interval $$\cos A > \sin A.$$

$$\eqalign{ & \sin x = \frac{{7 \pm \sqrt {49 - 24} }}{6} = \frac{{7 \pm 5}}{6} = \frac{1}{3},2\left( {{\text{absurd}}} \right). \cr & \therefore \,\,\sin x = \frac{1}{3} = \sin \alpha ,\,\,\,\,\,{\text{where }}0 < \alpha < \frac{\pi }{2}. \cr & \therefore \,\,x = \alpha ,\pi - \alpha ,2\pi + \alpha ,3\pi - \alpha ,4\pi + \alpha ,5\pi - \alpha . \cr} $$

The given expression is $$\sqrt 3 \,{\text{cosec}}\,{\text{2}}{{\text{0}}^ \circ } - \sec {20^ \circ }$$
$$\eqalign{ & = \frac{{\sqrt 3 }}{{\sin {{20}^ \circ }}} - \frac{1}{{\cos {{20}^ \circ }}} = \frac{{\sqrt 3 \cos {{20}^ \circ } - \sin {{20}^ \circ }}}{{\sin {{20}^ \circ }\cos {{20}^ \circ }}} \cr & = 4\left[ {\frac{{\frac{{\sqrt 3 }}{2}\cos {{20}^ \circ } - \frac{1}{2}\sin {{20}^ \circ }}}{{2\sin {{20}^ \circ }\cos {{20}^ \circ }}}} \right] \cr & = 4\left[ {\frac{{\sin {{60}^ \circ }\cos {{20}^ \circ } - \cos {{60}^ \circ }\sin {{20}^ \circ }}}{{\sin \left( {2 \times {{20}^ \circ }} \right)}}} \right] \cr & = \frac{{4\sin \left( {{{60}^ \circ } - {{20}^ \circ }} \right)}}{{\sin {{40}^ \circ }}} \cr & = \frac{{4\sin {{40}^ \circ }}}{{\sin {{40}^ \circ }}} \cr & = 4 \cr} $$

$$\eqalign{ & \cos \left( {\alpha + \beta } \right) = \frac{4}{5} \cr & \Rightarrow \,\,\tan \left( {\alpha + \beta } \right) = \frac{3}{4} \cr & \sin \left( {\alpha - \beta } \right) = \frac{5}{{13}} \cr & \Rightarrow \,\,\tan \left( {\alpha - \beta } \right) = \frac{5}{{12}} \cr & \tan 2\alpha = \tan \left[ {\left( {\alpha + \beta } \right) + \left( {\alpha - \beta } \right)} \right] \cr & = \frac{{\frac{3}{4} + \frac{5}{{12}}}}{{1 - \frac{3}{4}.\frac{5}{{12}}}} \cr & = \frac{{56}}{{33}} \cr} $$

Value $$ = 1 + \frac{1}{{\sqrt 2 }}\left( {\cos \theta + \sin \theta } \right) + \sqrt 2 \left( {\cos \theta + \sin \theta } \right)$$
$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 + \left( {\frac{1}{{\sqrt 2 }} + \sqrt 2 } \right) \cdot \left( {\cos \theta + \sin \theta } \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 + \left( {\frac{1}{{\sqrt 2 }} + \sqrt 2 } \right) \cdot \sqrt 2 \cos \left( {\theta - \frac{\pi }{4}} \right). \cr} $$
∴ the maximum value $$ = 1 + \left( {\frac{1}{{\sqrt 2 }} + \sqrt 2 } \right)\sqrt 2 = 4.$$

$$2\sin \alpha \cdot \cos \beta \cdot \sin \gamma = \sin \alpha \cdot \sin \beta \cdot \cos\gamma + \sin\beta \cdot \sin\gamma \cdot \cos\alpha .$$
Dividing by $$\sin\alpha \cdot \sin\beta \cdot \sin\gamma ,2\cot\beta = \cot\gamma + \cot\alpha .$$

$$\eqalign{ & {\left( {{e^{\sin x}}} \right)^2} - 4{e^{\sin x}} - 1 = 0 \cr & \therefore \,\,{e^{\sin x}} = \frac{{4 \pm \sqrt {16 + 4} }}{2} = 2 \pm \sqrt 5 , \cr} $$
which is absurd because $$0 < {e^{\sin x}} \leqslant e = 2.72\left( {{\text{nearly}}} \right).$$