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

$$\eqalign{ & {\cos ^2}{10^ \circ } - \cos {10^ \circ }\cos {50^ \circ } + {\cos ^2}{50^ \circ } \cr & = \left( {\frac{{1 + \cos {{20}^ \circ }}}{2}} \right) + \left( {\frac{{1 + \cos {{100}^ \circ }}}{2}} \right) - \frac{1}{2}\left( {2\cos {{10}^ \circ }\cos {{50}^ \circ }} \right) \cr & = 1 + \frac{1}{2}\left( {\cos {{20}^ \circ } + \cos {{100}^ \circ }} \right) - \frac{1}{2}\left[ {\cos {{60}^ \circ } + \cos {{40}^ \circ }} \right] \cr & = \left( {1 - \frac{1}{4}} \right) + \frac{1}{2}\left[ {\cos {{20}^ \circ } + \cos {{100}^ \circ } - \cos {{40}^ \circ }} \right] \cr & = \frac{3}{4} + \frac{1}{2}\left[ {2\cos {{60}^ \circ } \times \cos {{40}^ \circ } - \cos {{40}^ \circ }} \right] \cr & = \frac{3}{4} \cr} $$

$$\eqalign{ & {\left( {1 - \sin A + \cos A} \right)^2} \cr & = 1 + {\sin ^2}A + {\cos ^2}A - 2\sin A - 2\sin A \cdot \cos A + 2\cos A \cr & = 2 - 2\sin A - 2\sin A\cos A + 2\cos A \cr & = 2\left( {1 - \sin A} \right) + 2\cos A\left( {1 - \sin A} \right) \cr & = 2\left( {1 + \cos A} \right)\left( {1 - \sin A} \right) \cr} $$

For checking equal function
$$\left( A \right)$$  Domain of $$f\left( x \right) = R$$   but range $$ = \left[ {0,\infty } \right)$$
Domain of $$g\left( x \right) = R,$$   range $$ = R$$
Domain same but range is different so it is not an equal function.
$$\left( B \right)$$  Domain of $$f\left( x \right) = R$$
Domain of $$g\left( x \right) = R$$
Domain and range both same so it is an equal function.
$$\left( C \right)$$  Domain of $$f\left( x \right) = R - \left\{ 0 \right\}$$
Domain of $$g\left( x \right) = R$$
Not equal function as domain is different.

$$\eqalign{ & \tan \left( {A + B + C} \right) = 0\,\left( {{\text{from the question}}} \right) \cr & \therefore \,\,A + B + C = n\pi . \cr} $$

$$\eqalign{ & \alpha + \beta + \gamma = 2\pi \cr & \Rightarrow \,\,\frac{\alpha }{2} + \frac{\beta }{2} + \frac{\gamma }{2} = \pi \cr & \therefore \,\,\tan \left( {\frac{\alpha }{2} + \frac{\beta }{2}} \right) = \tan \left( {\pi - \frac{\gamma }{2}} \right) = - \tan \frac{\gamma }{2} \cr & \Rightarrow \,\,\frac{{\tan \frac{\alpha }{2} + \tan \frac{\beta }{2}}}{{1 - \tan \frac{\alpha }{2}\tan \frac{\beta }{2}}} = - \tan \frac{\gamma }{2} \cr & \Rightarrow \,\,\tan \frac{\alpha }{2} + \tan \frac{\beta }{2} + \tan \frac{\gamma }{2} = \tan \frac{\alpha }{2}\tan \frac{\beta }{2}\tan \frac{\gamma }{2} \cr} $$

$$\eqalign{ & \cos \alpha = \frac{1}{2}\left( {x + \frac{1}{x}} \right) \cr & \Rightarrow \,\,x = \cos \alpha \pm i\sin \alpha . \cr & {\text{Similarly, }}y = \cos \beta \pm i\sin \beta . \cr & \therefore \,\,\frac{x}{y} = \cos \left( {\alpha - \beta } \right) \pm i\sin \left( {\alpha - \beta } \right), \cr & xy = \cos \left( {\alpha + \beta } \right) \pm i\sin \left( {\alpha + \beta } \right) \cr & {\text{and, }}\frac{y}{x} = \cos \left( {\alpha - \beta } \right) \mp i\sin \left( {\alpha - \beta } \right). \cr} $$

Given $$3\sin P + 4\cos Q = 6\,\,\,\,.....\left( {\text{i}} \right)$$
$$4\sin Q + 3\cos P = 1\,\,\,\,.....\left( {{\text{ii}}} \right)$$
Squaring and adding (i) & (ii) we get
$$\eqalign{ & 9{\sin ^2}P + 16{\cos ^2}Q + 24\sin P\cos Q + 16{\sin ^2}Q + 9{\cos ^2}P + 24\sin Q\cos P = 36 + 1 = 37 \cr & \Rightarrow \,\,9\left( {{{\sin }^2}P + {{\cos }^2}P} \right) + 16\left( {{{\sin }^2}Q + {{\cos }^2}Q} \right) + 24\left( {\sin P\cos Q + \cos P\sin Q} \right) = 37 \cr & \Rightarrow \,\,9 + 16 + 24\sin \left( {P + Q} \right) = 37 \cr & \left[ {\because \,\,{{\sin }^2}\theta + {{\cos }^2}\theta = 1\,\,{\text{and }}\sin A\cos B + \cos A\sin B = \sin \left( {A + B} \right)} \right] \cr & \Rightarrow \,\,\sin \left( {P + Q} \right) = \frac{1}{2} \cr & \Rightarrow \,\,P + Q = \frac{\pi }{6}\,\,{\text{or }}\frac{{5\pi }}{6} \cr & \Rightarrow \,\,R = \frac{{5\pi }}{6}\,\,{\text{or }}\frac{\pi }{6}\,\,\left( {\because \,\,P + Q + R = \pi } \right) \cr & {\text{If }}R = \frac{{5\pi }}{6}\,\,{\text{then 0}} < P,Q < \frac{\pi }{6} \cr & \Rightarrow \,\,\cos Q < 1\,\,{\text{and }}\sin P < \frac{1}{2} \cr} $$
$$ \Rightarrow \,\,3\sin P + 4\cos Q < \frac{{11}}{2}$$       which is not true.
$${\text{So, }}R = \frac{\pi }{6}$$

We are given that
$$\eqalign{ & \left( {\cot \,{\alpha _1}} \right).\left( {\cot \,{\alpha _2}} \right)\,.....\,\left( {\cot \,{\alpha _n}} \right) = 1 \cr & \Rightarrow \,\left( {\cos \,{\alpha _1}} \right)\left( {\cos \,{\alpha _2}} \right)\,.....\,\left( {\cos \,{\alpha _n}} \right) \cr & = \left( {\sin \,{\alpha _1}} \right)\,\left( {\sin \,{\alpha _2}} \right)\,.....\left( {\sin \,{\alpha _n}} \right)\,\,\,\,\,\,\,\,\,.....\left( 1 \right) \cr & {\text{Let}}\,y\,{\text{ = }}\left( {\cos \,{\alpha _1}} \right)\,\left( {\cos \,{\alpha _2}} \right)....\,\left( {\cos \,{\alpha _n}} \right)\,\left( {{\text{to}}\,{\text{be}}\,{\text{max}}{\text{.}}} \right) \cr & {\text{Squaring}}\,{\text{both}}\,{\text{sides,}}\,{\text{we}}\,{\text{get}} \cr & {y^{\text{2}}}{\text{ = }}\,\left( {{{\cos }^2}\,{\alpha _1}} \right)\,\,\left( {{{\cos }^2}\,{\alpha _2}} \right)\,.....\,\left( {{{\cos }^2}\,{\alpha _n}} \right) \cr & = \,\cos \,{\alpha _1}\,\sin \,{\alpha _1}\,\cos \,{\alpha _2}\,\sin \,{\alpha _2}\,.....\,\cos \,{\alpha _n}\,\sin \,{\alpha _n}\,\,\,\,\left( {{\text{Using}}\,\left( 1 \right)} \right) \cr & = \,\frac{1}{{{2^n}}}\,\left[ {\sin \,2{\alpha _1}\,\sin \,2{\alpha _2}\,.....\sin 2\,{\alpha _n}} \right] \cr & {\text{As}}\,\,0 \leqslant {\alpha _1},\,{\alpha _2},.....{\alpha _n} \leqslant \frac{\pi }{2} \cr & \therefore \,\,\,0 \leqslant 2{\alpha _1},\,2{\alpha _2},.....\sin \,2{\alpha _n}\, \leqslant \pi \cr & \Rightarrow \,\,0 \leqslant \,\sin \,2{\alpha _1},\,\sin \,2{\alpha _2},.....\,\sin \,2{\alpha _n}\, \leqslant 1 \cr & \therefore \,{y^2}\, \leqslant \,\frac{1}{{{2^n}}}.1 \cr & \Rightarrow \,\,y \leqslant \,\frac{1}{{{2^{\frac{n}{2}}}}} \cr} $$
∴ Max. value of $$y$$ is $$\frac{1}{{{2^{\frac{n}{2}}}}}.$$

Differentiate w.r.t. $$\theta $$ and put $$\theta = \frac{\pi }{2}.$$

$$\eqalign{ & {x^2} - \cos \theta \cdot x + p = 0. \cr & \therefore \,\,x = \frac{{\cos \theta \pm \sqrt {{{\cos }^2}\theta - 4p} }}{2}. \cr} $$
For real $$x,$$
$$\eqalign{ & {\cos ^2}\theta - 4p \geqslant 0. \cr & \therefore \,\,4p \leqslant {\cos ^2}\theta \leqslant 1. \cr} $$