291.
Let $${\text{a}} = \frac{{2\sin x}}{{1 + \sin x + \cos x}}$$ ÃÂÃÂ ÃÂÃÂ and $${\text{b}} = \frac{{\text{c}}}{{1 + \sin x}}.$$ ÃÂÃÂ Then a = b, if c = ?
(A) 1 - sinxcosx
(B) 1 + sinx - cosx
(C) 1 + sinxcosx
(D) 1 + cosx - sinx
Show Answer
Solution:
$$\eqalign{ & b = \frac{c}{{1 + \sin x}} \cr & {\text{Go through option from option B}} \cr & b = \frac{{1 + \sin x - \cos x}}{{1 + \sin x}} \cr & = \frac{{{{\left( {1 + \sin x} \right)}^2} - {{\cos }^2}x}}{{1 + \sin x\left( {1 + \sin x + \cos x} \right)}} \cr & = \frac{{1 + {{\sin }^2}x + 2{{\sin }^2}x - 1 + {{\sin }^2}x}}{{\left( {1 + \sin x} \right)\left( {1 + \sin x + \cos x} \right)}} \cr & = \frac{{2{{\sin }^2}x + 2\sin x}}{{\left( {1 + \sin x} \right)\left( {1 + \sin x + \cos x} \right)}} \cr & = \frac{{2\sin x\left( {\sin x + 1} \right)}}{{1 + \sin x + \cos x}} \cr & = \frac{{2\sin x}}{{1 + \sin x + \cos x}} \cr & = a \cr} $$
292.
What will be the value of sin10ÃÂÃÂÃÂð - $$\frac{4}{3}$$sin3 10ÃÂÃÂÃÂð?
(A) $$\frac{1}{{3\sqrt 3 }}$$
(B) $$\frac{1}{6}$$
(C) $$\frac{1}{{2\sqrt 3 }}$$
(D) $$\frac{{\sqrt 3 }}{6}$$
Show Answer
Solution:
$$\eqalign{ & \sin 10 - \frac{4}{3}{\sin ^3}10 \cr & = \frac{{3\sin 10 - 4{{\sin }^3}10}}{3} \cr & = \frac{1}{3}\sin 3 \times {10^ \circ }\,\,\,\,\,\left[ {\because \sin A = 3\sin A - 4{{\sin }^3}A} \right] \cr & = \frac{1}{3} \times \frac{1}{2} \cr & = \frac{1}{6} \cr} $$
293.
What is the value of $$\frac{{{{\left[ {1 - \tan \left( {{{90}^ \circ } - \theta } \right)} \right]}^2}}}{{\left[ {{{\cos }^2}\left( {{{90}^ \circ } - \theta } \right)} \right]}} - 1 = ?$$
(A) -sin2θ
(B) -cos2θ
(C) cos2θ
(D) sin2θ
Show Answer
Solution:
$$\eqalign{ & \frac{{{{\left[ {1 - \tan \left( {{{90}^ \circ } - \theta } \right)} \right]}^2}}}{{\left[ {{{\cos }^2}\left( {{{90}^ \circ } - \theta } \right)} \right]}} - 1 \cr & {\text{By putting }}\theta = {45^ \circ } \cr & \Rightarrow \frac{{{{\left[ {1 - \tan \left( {{{90}^ \circ } - {{45}^ \circ }} \right)} \right]}^2}}}{{\left[ {{{\cos }^2}\left( {{{90}^ \circ } - {{45}^ \circ }} \right)} \right]}} - 1 \cr & \Rightarrow \frac{{{{\left[ {1 - \tan {{45}^ \circ }} \right]}^2}}}{{{{\cos }^2}{{45}^ \circ }}} - 1 \cr & \Rightarrow 0 - 1 \cr & \Rightarrow - 1 \cr & {\text{By satisfying in option A}} \cr & \Rightarrow - \sin 2\theta \cr & \Rightarrow - \sin {90^ \circ } \cr & \Rightarrow - 1 \cr} $$
294.
The value of cos2 20ÃÂÃÂÃÂð + cos2 70ÃÂÃÂÃÂð is?
(A) $$\sqrt 2 $$
(B) 2
(C) $$\frac{1}{{\sqrt 2 }}$$
(D) 1
Show Answer
Solution:
$$\eqalign{ & {\text{co}}{{\text{s}}^2}{20^ \circ } + {\text{co}}{{\text{s}}^2}{70^ \circ } \cr & = {\text{co}}{{\text{s}}^2}\left( {{{90}^ \circ } - {{70}^ \circ }} \right) + {\text{co}}{{\text{s}}^2}{70^ \circ } \cr & = {\sin ^2}{70^ \circ } + {\text{co}}{{\text{s}}^2}{70^ \circ } \cr & = 1 \cr} $$
295.
What is the value of $$1 + \frac{{{{\tan }^2}A}}{{1 + \sec A}}?$$
(A) cosecA
(B) cosA
(C) secA
(D) sinA
Show Answer
Solution:
$$\eqalign{ & 1 + \frac{{{{\tan }^2}A}}{{1 + \sec A}} \cr & = 1 + \frac{{{{\sec }^2}A - 1}}{{1 + \sec A}} \cr & = \frac{{1 + \sec A + {{\sec }^2}A - 1}}{{1 + \sec A}} \cr & = \frac{{\sec A\left( {1 + \sec A} \right)}}{{1 + \sec A}} \cr & = \sec A \cr} $$
296.
If secA = $$\frac{{17}}{8},$$ given that A 90ÃÂÃÂÃÂð, what is the value of the following?
(A) 23
(B) 19
(C) 30
(D) 38
Show Answer
Solution:
$$\eqalign{ & \sec A = \frac{{17 \to H}}{{8 \to B}},\,A {90^ \circ } \cr & {L^2} = {H^2} - {B^2} \cr & {L^2} = 289 - 64 \cr & {L^2} = 225 \cr & L = 15 \cr & {\text{Then}}, \cr & \frac{{34\sin A + 15\cot A}}{{68\cos A - 16\tan A}} \cr & = \frac{{34 \times \frac{{15}}{{17}} + 15 \times \frac{8}{{15}}}}{{68 \times \frac{8}{{17}} - 16 \times \frac{{15}}{8}}} \cr & = \frac{{30 + 8}}{{32 - 30}} \cr & = \frac{{38}}{2} \cr & = 19 \cr} $$
297.
If A = sin2 ÃÂÃÂÃÂø + cos4 ÃÂÃÂÃÂø for any value of ÃÂÃÂÃÂø, then the value of A is?
(A) $${\text{1}} \leqslant {\text{A}} \leqslant {\text{1}}$$
(B) $$\frac{3}{4} \leqslant {\text{A}} \leqslant {\text{1}}$$
(C) $$\frac{{13}}{{16}} \leqslant {\text{A}} \leqslant {\text{1}}$$
(D) $$\frac{3}{4} \leqslant {\text{A}} \leqslant \frac{{13}}{{16}}$$
Show Answer
Solution:
$$\eqalign{ & {\text{According to the question, }} \cr & {\text{A}} = {\sin ^2}\theta + {\text{co}}{{\text{s}}^4}\theta \cr & {\text{Put }}\theta = {90^ \circ }{\text{ for maximum value of A}} \cr & {\text{A}} = {\sin ^2}{90^ \circ } + {\text{co}}{{\text{s}}^4}{90^ \circ } \cr & {\text{A}} = 1 + 0 \cr & {\text{A}} = 1 \cr & {\text{Put }}\theta = {45^ \circ }{\text{ for minimum value of A}} \cr & {\text{A}} = {\sin ^2}{45^ \circ } + {\text{co}}{{\text{s}}^4}{45^ \circ } \cr & {\text{A}} = \frac{1}{2} + \frac{1}{4} \cr & {\text{A}} = \frac{3}{4} \cr & \therefore {\text{A lies in }}\frac{3}{4} \leqslant {\text{A}} \leqslant {\text{1}} \cr} $$
298.
If cos20ÃÂÃÂÃÂð = m and cos70ÃÂÃÂÃÂð =n, then the value of m2 + n2 is?
(A) $$\frac{1}{2}$$
(B) 1
(C) $$\frac{3}{2}$$
(D) $$\frac{1}{{\sqrt 2 }}$$
Show Answer
Solution:
$$\eqalign{ & \cos {20^ \circ } = m{\text{ }} \cr & \cos {70^ \circ } = n \cr & {\text{So,}} \cr & \Leftrightarrow {m^2} + {n^2} = {\text{co}}{{\text{s}}^2}{20^ \circ } + {\text{co}}{{\text{s}}^2}{70^ \circ } \cr & \left[ {{\text{If co}}{{\text{s}}^2}{\text{A + co}}{{\text{s}}^2}{\text{B}} = {\text{1}}} \right] \cr & ({\text{If, A}} + {\text{B}} = {90^ \circ }) \cr & \Leftrightarrow 1 \cr} $$
299.
The value of following is, cos24ÃÂÃÂÃÂð + cos55ÃÂÃÂÃÂð + cos125ÃÂÃÂÃÂð + cos204ÃÂÃÂÃÂð + cos300ÃÂÃÂÃÂð ?
(A) $$ - \frac{1}{2}$$
(B) $$\frac{1}{2}$$
(C) 2
(D) 1
Show Answer
Solution:
The value of, cos24° + cos55° + cos125° + cos204° + cos300° We know that, cos(180° $$ \pm $$ θ) = -cosθ ⇒ cos24° + cos55° + cos (180° - 55°) + cos (180° + 24°) + cos (360° - 60°) ⇒ cos24° + cos55° - cos55° - cos24° + cos60° ⇒ cos60° ⇒ $$\frac{1}{2}$$
300.
What is the value of $$\frac{{\cos {{40}^ \circ } - \cos {{140}^ \circ }}}{{\sin {{80}^ \circ } + \sin {{20}^ \circ }}}?$$
(A) $$2\sqrt 3 $$
(B) $$\frac{2}{{\sqrt 3 }}$$
(C) $$\frac{1}{{\sqrt 3 }}$$
(D) $$\sqrt 3 $$
Show Answer
Solution:
$$\eqalign{ & \frac{{\cos {{40}^ \circ } - \cos {{140}^ \circ }}}{{\sin {{80}^ \circ } + \sin {{20}^ \circ }}} \cr & = \frac{{ - 2\sin \left( {\frac{{{{40}^ \circ } + {{140}^ \circ }}}{2}} \right)\sin \left( {\frac{{{{40}^ \circ } - {{140}^ \circ }}}{2}} \right)}}{{2\sin \left( {\frac{{{{80}^ \circ } + {{20}^ \circ }}}{2}} \right)\cos \left( {\frac{{{{80}^ \circ } - {{20}^ \circ }}}{2}} \right)}} \cr & = \frac{{ - 2\sin \left( {\frac{{{{180}^ \circ }}}{2}} \right)\sin \left( { - \frac{{{{100}^ \circ }}}{2}} \right)}}{{2\sin \left( {\frac{{{{100}^ \circ }}}{2}} \right)\cos \left( {\frac{{{{60}^ \circ }}}{2}} \right)}} \cr & = \frac{{ - 2\sin {{90}^ \circ } \times \sin \left( { - {{50}^ \circ }} \right)}}{{2\sin {{50}^ \circ } \times \cos {{30}^ \circ }}} \cr & = \frac{{ - 2\sin {{90}^ \circ } \times \left( { - \sin {{50}^ \circ }} \right)}}{{2\sin {{50}^ \circ } \times \cos {{30}^ \circ }}} \cr & = \frac{{2\sin {{90}^ \circ } \times \sin {{50}^ \circ }}}{{2\sin {{50}^ \circ } \times \cos {{30}^ \circ }}} \cr & = \frac{{\sin {{90}^ \circ }}}{{\cos {{30}^ \circ }}} \cr & = \frac{1}{{\frac{{\sqrt 3 }}{2}}} \cr & = \frac{2}{{\sqrt 3 }} \cr} $$