91.
If tan ÃÂÃÂÃÂø = 1, then the value of $$\frac{{8\sin \theta + 5\cos \theta }}{{{{\sin }^3}\theta - 2{{\cos }^3}\theta + 7\cos \theta }}$$ ÃÂàÃÂàis?
(A) 2
(B) $${\text{2}}\frac{1}{2}$$
(C) 3
(D) $$\frac{4}{5}$$
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Solution:
$$\eqalign{ & {\text{If }}\tan \theta = 1 \cr & {\text{It means }}\theta = {45^ \circ } \cr & = \frac{{8\sin \theta + 5\cos \theta }}{{{{\sin }^3}\theta - 2{{\cos }^3}\theta + 7\cos \theta }} \cr & = \frac{{8\sin {{45}^ \circ } + 5\cos {{45}^ \circ }}}{{{{\sin }^3}{{45}^ \circ } - 2{{\cos }^3}{{45}^ \circ } + 7\cos {{45}^ \circ }}} \cr & = \frac{{8 \times \frac{1}{{\sqrt 2 }} + 5 \times \frac{1}{{\sqrt 2 }}}}{{{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^3} - 2{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^3} + 7\left( {\frac{1}{{\sqrt 2 }}} \right)}} \cr & = 2 \cr} $$
92.
If cos53ÃÂÃÂÃÂð = $$\frac{x}{y},$$ then sec53ÃÂÃÂÃÂð + cot37ÃÂÃÂÃÂð is equal to:
(A) $$\frac{{x + \sqrt {{y^2} - {x^2}} }}{y}$$
(B) $$\frac{{x + \sqrt {{y^2} - {x^2}} }}{x}$$
(C) $$\frac{{y + \sqrt {{y^2} - {x^2}} }}{x}$$
(D) $$\frac{{y + \sqrt {{y^2} - {x^2}} }}{y}$$
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Solution:
$$\eqalign{ & \cot {53^ \circ } = \frac{x}{y} = \frac{B}{H} \cr & {P^2} = {H^2} - {B^2} \cr & P = \sqrt {{y^2} - {x^2}} \cr & \sec {53^ \circ } + \cot {37^ \circ } \cr & = \sec {53^ \circ } + \tan {53^ \circ } \cr & = \frac{H}{B} + \frac{P}{B} \cr & = \frac{{H + P}}{B} \cr & = \frac{{y + \sqrt {{y^2} - {x^2}} }}{x} \cr} $$
93.
If ÃÂÃÂÃÂø be acute angle and tan(4ÃÂÃÂÃÂø - 50ÃÂÃÂÃÂð) = cot(50ÃÂÃÂÃÂð - ÃÂÃÂÃÂø), then the value of ÃÂÃÂÃÂø in degrees is?
(A) 30°
(B) 40°
(C) 50°
(D) 20°
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Solution:
$$\eqalign{ & {\text{We know that }} \cr & {\text{tan}}\left( {{{90}^ \circ } - \theta } \right) = {\text{cot}}\theta \cr & {\text{and, cot}}\left( {{{90}^ \circ } - \theta } \right) = {\text{tan}}\theta \cr & \Rightarrow {\text{tan}}\left( {4\theta - {{50}^ \circ }} \right) = {\text{cot}}\left( {{{50}^ \circ } - \theta } \right) \cr & \Rightarrow \cot \left[ {{{90}^ \circ } - \left( {4\theta - {{50}^ \circ }} \right)} \right] = {\text{cot}}\left( {{{50}^ \circ } - \theta } \right) \cr & \Rightarrow {90^ \circ } - \left( {4\theta - {{50}^ \circ }} \right) = \left( {{{50}^ \circ } - \theta } \right) \cr & \Rightarrow {90^ \circ } - 4\theta + {50^ \circ } = {50^ \circ } - \theta \cr & \Rightarrow {90^ \circ } = 3\theta \cr & {\text{then}},\theta = {30^ \circ } \cr} $$
94.
If $$\sin \left( {\theta + {{30}^ \circ }} \right) = \frac{3}{{\sqrt {12} }}{\text{,}}$$ ÃÂÃÂ ÃÂÃÂ then find $${\text{co}}{{\text{s}}^2}\theta ?$$
(A) $$\frac{1}{4}$$
(B) $$\frac{3}{4}$$
(C) $$\frac{{\sqrt 3 }}{2}$$
(D) $$\frac{1}{2}$$
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Solution:
$$\eqalign{ & \sin \left( {\theta + {{30}^ \circ }} \right) = \frac{3}{{\sqrt {12} }} = \frac{3}{{2\sqrt 3 }} \cr & = \frac{{\sqrt 3 }}{2} = \sin \left( {\theta + {{30}^ \circ }} \right) = \sin {60^ \circ } \cr & \therefore \theta = {30^ \circ } \cr & {\text{co}}{{\text{s}}^2}\theta = {\text{co}}{{\text{s}}^2}{30^ \circ } \cr & = {\left( {\frac{{\sqrt 3 }}{2}} \right)^2} \cr & = \frac{3}{4} \cr} $$
95.
What is the value of $$\frac{{\left[ {1 - \tan \left( {90 - \theta } \right) + \sec \left( {90 - \theta } \right)} \right]}}{{\left[ {\tan \left( {90 - \theta } \right) - \sec \left( {90 - \theta } \right) + 1} \right]}}?$$
(A) $$\cot \frac{\theta }{2}$$
(B) $$\tan \frac{\theta }{2}$$
(C) sinθ
(D) cosθ
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Solution:
\[\begin{array}{l} \frac{{\left[ {1 - \tan \left( {90 - \theta } \right) + \sec \left( {90 - \theta } \right)} \right]}}{{\left[ {\tan \left( {90 - \theta } \right) - \sec \left( {90 - \theta } \right) + 1} \right]}}\\ \Rightarrow \frac{{\left[ {1 - \cot \theta + {\rm{cosec}}\,\theta } \right]}}{{\left[ {\cot \theta + {\rm{cosec}}\,\theta + 1} \right]}}\\ \Rightarrow \frac{{\left[ {1 - \frac{{\cos \theta }}{{\sin \theta }} + \frac{1}{{\sin \theta }}} \right]}}{{\left[ {\frac{{\cos \theta }}{{\sin \theta }} + \frac{1}{{\sin \theta }} + 1} \right]}}\\ \Rightarrow \frac{{\left[ {\sin \theta - \cos \theta + 1} \right]}}{{\left[ {\sin \theta + \cos \theta + 1} \right]}}\\ \Rightarrow \frac{{\left( {\sin \theta + 1} \right) - \cos \theta }}{{\left( {\sin \theta + 1} \right) + \cos \theta }}\\ \left[ \begin{array}{l} \therefore \sin \theta = 2\sin \frac{\theta }{2}.\cos \frac{\theta }{2}\\ \cos \theta = 1 - 2{\sin ^2}\frac{\theta }{2}\\ \cos \frac{\theta }{2} = 2{\cos ^2}\frac{\theta }{2} - 1 \end{array} \right]\\ \Rightarrow \frac{{2\sin \frac{\theta }{2}.\cos \frac{\theta }{2} + 1 - 1 + 2{{\sin }^2}\frac{\theta }{2}}}{{2\sin \frac{\theta }{2}.\cos \frac{\theta }{2} + 1 + 2{{\cos }^2}\frac{\theta }{2} - 1}}\\ \Rightarrow \frac{{2\sin \frac{\theta }{2}\left( {\sin \frac{\theta }{2} + \cos \frac{\theta }{2}} \right)}}{{2\cos \frac{\theta }{2}\left( {\sin \frac{\theta }{2} + \cos \frac{\theta }{2}} \right)}}\\ \Rightarrow \tan \frac{\theta }{2} \end{array}\]
96.
The value of $${\left( {\frac{{1 - \cot \theta }}{{1 - \tan \theta }}} \right)^2} + 1,$$ ÃÂàÃÂà0ÃÂÃÂÃÂð ÃÂÃÂÃÂø 90ÃÂÃÂÃÂð, is equal to:
(A) cosec2θ
(B) sin2θ
(C) cos2θ
(D) sec2θ
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Solution:
$$\eqalign{ & {\left( {\frac{{1 - \cot \theta }}{{1 - \tan \theta }}} \right)^2} + 1 \cr & = {\left( {\frac{{1 - \frac{{\cos \theta }}{{\sin \theta }}}}{{1 + \frac{{\sin \theta }}{{\cos \theta }}}}} \right)^2} + 1 \cr & = {\left( {\frac{{\sin \theta - \cos \theta }}{{\sin \theta }} \times \frac{{\cos \theta }}{{\cos \theta - \sin \theta }}} \right)^2} + 1 \cr & = {\left( {\frac{{\sin \theta - \cos \theta \times \cos \theta }}{{ - \sin \theta \left( {\sin \theta - \cos \theta } \right)}}} \right)^2} + 1 \cr & = {\left( {\frac{{ - \cos \theta }}{{\sin \theta }}} \right)^2} + 1 \cr & = {\cot ^2}\theta + 1 \cr & = {\text{cose}}{{\text{c}}^2}\theta \cr} $$
97.
The value of the expression 2(sin6 ÃÂÃÂÃÂø + cos6 ÃÂÃÂÃÂø) - 3(sin4 ÃÂÃÂÃÂø + cos4 ÃÂÃÂÃÂø) + 1 is?
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Solution:
$${\text{ 2}}\left( {{{\sin }^6}\theta + {\text{co}}{{\text{s}}^6}\theta } \right) - {\text{3}}\left( {{{\sin }^4}\theta + {\text{co}}{{\text{s}}^4}\theta } \right){\text{ + 1}}$$ $$ \Rightarrow {\text{2}}\left( {1 - 3{{\sin }^2}\theta {\text{co}}{{\text{s}}^2}\theta } \right) - $$ $${\text{3}}\left( {1 - 2{{\sin }^2}\theta c{\text{o}}{{\text{s}}^2}\theta } \right){\text{ + }}$$ $${\text{1}}$$ $$\eqalign{ & \Rightarrow {\text{ 2}} - 6{\sin ^2}\theta .{\text{co}}{{\text{s}}^2}\theta - 3 + 6{\sin ^2}\theta .{\text{co}}{{\text{s}}^2}\theta {\text{ + 1}} \cr & \Rightarrow 2 - 3 + 1 \cr & \Rightarrow 0 \cr} $$
98.
If tanA = n tanB and sinA = m sinB, then the value of cos2 A = ?
(A) $$\frac{{{m^2} + 1}}{{{n^2} + 1}}$$
(B) $$\frac{{{m^2} + 1}}{{{n^2} - 1}}$$
(C) $$\frac{{{m^2} - 1}}{{{n^2} - 1}}$$
(D) $$\frac{{{m^2} - 1}}{{{n^2} + 1}}$$
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Solution:
$$\eqalign{ & {\text{sin A}} = m{\text{ sin B}} \cr & {\text{si}}{{\text{n}}^2}{\text{A}} = {m^2}{\text{si}}{{\text{n}}^2}{\text{B }}......{\text{(i)}} \cr & {\text{Now, ta}}{{\text{n}}^2}{\text{A}} = {n^2}{\text{ta}}{{\text{n}}^2}{\text{B}} \cr & \frac{{{{\sin }^2}{\text{A}}}}{{{\text{co}}{{\text{s}}^2}{\text{A}}}} = {n^2}\frac{{{{\sin }^2}{\text{B}}}}{{{\text{co}}{{\text{s}}^2}{\text{B}}}} \cr & {\text{from equation (i)}} \cr & \Rightarrow \frac{{1 - {\text{co}}{{\text{s}}^2}{\text{A}}}}{{{{\text{n}}^2}{\text{co}}{{\text{s}}^2}{\text{A}}}} = \frac{{{{\sin }^2}{\text{B}}}}{{{\text{co}}{{\text{s}}^2}{\text{B}}}} \cr & \Rightarrow \frac{{1 - {\text{co}}{{\text{s}}^2}{\text{A}}}}{{{{\text{n}}^2}{\text{co}}{{\text{s}}^2}{\text{A}}}} = \frac{{\frac{{\left( {1 - {\text{co}}{{\text{s}}^2}{\text{A}}} \right)}}{{{m^2}}}}}{{1 - \frac{{{{\sin }^2}{\text{A}}}}{{{m^2}}}}} \cr & \Rightarrow \frac{{1 - {\text{co}}{{\text{s}}^2}{\text{A}}}}{{{{\text{n}}^2}{\text{co}}{{\text{s}}^2}{\text{A}}}} = \frac{{{\text{1}} - {\text{co}}{{\text{s}}^2}{\text{A}}}}{{{m^2} - 1 + {\text{co}}{{\text{s}}^2}{\text{A}}}} \cr & \Rightarrow {m^2} - 1 + {\text{co}}{{\text{s}}^2}{\text{A}} = {{\text{n}}^2}{\text{co}}{{\text{s}}^2}{\text{A}} \cr & \Rightarrow {m^2} - 1 = {\text{co}}{{\text{s}}^2}\theta \left( {{n^2} - 1} \right) \cr & \Rightarrow {\text{co}}{{\text{s}}^2}{\text{A}} = \frac{{{m^2} - 1}}{{{n^2} - 1}} \cr} $$
99.
If 7sinÃÂÃÂÃÂÃÂÃÂÃÂÃÂø + 3cosÃÂÃÂÃÂÃÂÃÂÃÂÃÂø = 4, (0ÃÂÃÂÃÂÃÂÃÂÃÂÃÂð ÃÂÃÂÃÂâÃÂÃÂÃÂÃÂÃÂä ÃÂÃÂÃÂÃÂÃÂÃÂÃÂø ÃÂÃÂÃÂâÃÂÃÂÃÂÃÂÃÂä 90ÃÂÃÂÃÂÃÂÃÂÃÂÃÂð), then the value of ÃÂÃÂÃÂÃÂÃÂÃÂÃÂø is?22
(A) $$\frac{\pi }{2}$$
(B) $$\frac{\pi }{3}$$
(C) $$\frac{\pi }{6}$$
(D) $$\frac{\pi }{4}$$
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Solution:
$$\eqalign{ & {\text{7}}{\sin ^2}\theta + 3{\text{co}}{{\text{s}}^2}\theta = 4 \cr & \Rightarrow {\text{7}}{\sin ^2}\theta + 3\left( {{\text{1}} - {\text{si}}{{\text{n}}^2}\theta } \right) = 4 \cr & \Rightarrow {\text{7}}{\sin ^2}\theta + 3 - 3{\sin ^2}\theta = 4 \cr & \Rightarrow 4{\sin ^2}\theta = 1 \cr & \Rightarrow {\sin ^2}\theta = \frac{1}{4} \cr & \Rightarrow \sin \theta = \frac{1}{2} = {\text{sin 3}}{0^ \circ } \cr & \theta = {30^ \circ } = \frac{\pi }{6}\left[ {\because {\pi ^c} = {{180}^ \circ }} \right] \cr} $$
100.
The expression of $$\frac{{\cot \theta + \operatorname{cosec} \theta - 1}}{{\cot \theta + \operatorname{cosec} \theta + 1}}$$ ÃÂÃÂ ÃÂÃÂ is equal to?
(A) $$\frac{{1 + \cos \theta }}{{\sin \theta }}$$
(B) $$\frac{{1 - \cos \theta }}{{\sin \theta }}$$
(C) $$\frac{{\cot \theta + 1}}{{\operatorname{cosec} \theta }}$$
(D) $$\frac{{\cot \theta - 1}}{{\sin \theta }}$$
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Solution:
$$\eqalign{ & \frac{{\cot \theta + \operatorname{cosec} \theta - 1}}{{\cot \theta + \operatorname{cosec} \theta + 1}} \cr & {\text{Put }}\theta = {45^ \circ } \cr & = \frac{{1 + \sqrt 2 - 1}}{{1 + \sqrt 2 + 1}} \cr & = \frac{{\sqrt 2 }}{{2 + \sqrt 2 }} \cr & = \frac{{\sqrt 2 }}{{\sqrt 2 \left( {\sqrt 2 + 1} \right)}} \cr & = \frac{1}{{\sqrt 2 + 1}} \cr & = \sqrt 2 - 1 \cr & {\text{Now option B}} \cr & \frac{{1 - \cos \theta }}{{\sin \theta }} \cr & = \frac{{1 - \frac{1}{{\sqrt 2 }}}}{{\frac{1}{{\sqrt 2 }}}} \cr & = \sqrt 2 - 1{\text{ }}\left( {{\text{Satisfy}}} \right) \cr} $$