61.
If $$\theta $$ be acute angle and $$\cos \theta = \frac{{15}}{{17}}{\text{,}}$$ ÃÂÃÂ then the value of $${\text{cot}}\left( {{{90}^ \circ } - \theta } \right)$$ ÃÂÃÂ is?
(A) $$\frac{{2\sqrt 8 }}{{15}}$$
(B) $$\frac{8}{{15}}$$
(C) $$\frac{{\sqrt 2 }}{{17}}$$
(D) $$\frac{{8\sqrt 2 }}{{17}}$$
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Solution:
$$\cos \theta = \frac{{15 \to {\text{Base}}}}{{17 \to {\text{Hypo}}}}$$ $$\eqalign{ & {\text{Perpendicular = 8}} \cr & \Rightarrow {\text{cot}}\left( {{{90}^ \circ } - \theta } \right) \cr & \Rightarrow {\text{tan}}\theta = \frac{8}{{15}}\left[ {\therefore \tan \theta = \frac{{\text{P}}}{{\text{B}}}} \right] \cr} $$
62.
The value of 4sin2 30ÃÂÃÂÃÂð + 3cot2 60ÃÂÃÂÃÂð is:
(A) 1
(B) $$\frac{1}{{\sqrt 2 }}$$
(C) 2
(D) 0
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Solution:
$$\eqalign{ & 4{\sin ^2}{30^ \circ } + 3{\cot ^2}{60^ \circ } \cr & = 4 \times {\left( {\frac{1}{2}} \right)^2} + 3{\left( {\frac{1}{{\sqrt 3 }}} \right)^2} \cr & = 4 \times \frac{1}{4} + 3 \times \frac{1}{3} \cr & = 1 + 1 \cr & = 2 \cr} $$
63.
If $${\text{A}} \times {\text{tan}}\left( {\theta + {{150}^ \circ }} \right)$$ ÃÂÃÂ ÃÂÃÂ = $${\text{B}} \times \tan $$ $$\left( {\theta - {{60}^ \circ }} \right){\text{,}}$$ ÃÂÃÂ the value of $$\frac{{{\text{A}} - {\text{B}}}}{{{\text{A}} + {\text{B}}}}$$ ÃÂÃÂ is?
(A) $$ - \frac{{\sin \theta }}{2}$$
(B) $$\frac{{\sin 2\theta }}{2}$$
(C) $$\frac{{\cos 2\theta }}{2}$$
(D) 0
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Solution:
$$\eqalign{ & {\text{A}} \times {\text{tan}}\left( {\theta + {{150}^ \circ }} \right) = {\text{B}} \times \tan \left( {\theta - {{60}^ \circ }} \right) \cr & \frac{{\text{A}}}{{\text{B}}} = \frac{{\tan \left( {\theta - {{60}^ \circ }} \right)}}{{\tan \left( {\theta + {{150}^ \circ }} \right)}} \cr & {\text{Put }}\theta = {90^ \circ } \cr & \frac{{\text{A}}}{{\text{B}}} = \frac{{\tan \left( {{{90}^ \circ } - {{60}^ \circ }} \right)}}{{\tan \left( {{{90}^ \circ } + {{150}^ \circ }} \right)}} \cr & \frac{{\text{A}}}{{\text{B}}} = \frac{{\tan {{30}^ \circ }}}{{\tan \left( {{{180}^ \circ } + {{60}^ \circ }} \right)}} \cr & \frac{{\text{A}}}{{\text{B}}} = \frac{{\tan {{30}^ \circ }}}{{\tan {{60}^ \circ }}} \cr & \frac{{\text{A}}}{{\text{B}}} = \frac{1}{3} \cr & {\text{then, }}\frac{{{\text{A}} + {\text{B}}}}{{{\text{A}} - {\text{B}}}} = - \frac{4}{2} \cr & \Rightarrow \frac{{{\text{A}} + {\text{B}}}}{{{\text{A}} - {\text{B}}}} = - 2 \cr & \Rightarrow \frac{{{\text{A}} - {\text{B}}}}{{{\text{A}} + {\text{B}}}} = - \frac{1}{2} \cr & {\text{Put in option (i)}} \cr & - \frac{{\sin {{90}^ \circ }}}{2} = - \frac{1}{2} \cr & {\text{So, option (A) is correct }} \cr} $$
64.
If A + B = C, then tanAtanBtanC = ?
(A) tanC + tanA - tanB
(B) tanC + tanA + tanB
(C) tanA - tanB - tanC
(D) tanC - tanA - tanB
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Solution:
A + B = C tan(A + B) = tanC $$\frac{{\tan {\text{A}} + \tan {\text{B}}}}{{1 - \tan {\text{A}}\tan {\text{B}}}} = \tan {\text{C}}$$ tanA + tanB = tanC - tanAtanBtanC tanAtanBtanC = tanC - tanA - tanB
65.
If sec(4x - 50ÃÂÃÂÃÂð) = cosec(50ÃÂÃÂÃÂð - x), then the value of x is?
(A) 45°
(B) 90°
(C) 30°
(D) 60°
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Solution:
$$\eqalign{ & {\text{sec}}\left( {4x - {{50}^ \circ }} \right) = {\text{cosec}}\left( {{{50}^ \circ } - x} \right) \cr & {\text{sec}}\left( {4x - {{50}^ \circ }} \right) = {\text{cosec}}\left( {{{90}^ \circ } - \left( {{{40}^ \circ } + x} \right)} \right) \cr & {\text{sec}}\left( {4x - {{50}^ \circ }} \right) = {\text{sec}}\left( {{{40}^ \circ } + x} \right) \cr & 4x - {50^ \circ } = {40^ \circ } + x \cr & 3x = {90^ \circ } \cr & x = {30^ \circ } \cr} $$
66.
The value of $$\frac{{\sin {{25}^ \circ }.\cos {{65}^ \circ } + \cos {{25}^ \circ }.\sin {{65}^ \circ }}}{{{{\tan }^2}{{70}^ \circ } - {\text{cose}}{{\text{c}}^2}{{20}^ \circ }}}$$ ÃÂÃÂ ÃÂÃÂ ÃÂÃÂ is?
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Solution:
$$\frac{{\sin {{25}^ \circ }.\cos {{65}^ \circ } + \cos {{25}^ \circ }.\sin {{65}^ \circ }}}{{{{\tan }^2}{{70}^ \circ } - \cos e{c^2}{{20}^ \circ }}}$$ $$ = \frac{{\sin {{25}^ \circ }.\cos \left( {{{90}^ \circ } - {{25}^ \circ }} \right) + \,\cos {{25}^ \circ }.\,\sin \left( {{{90}^ \circ } - {{25}^ \circ }} \right)}}{{{{\tan }^2}{{70}^ \circ } - {\text{cose}}{{\text{c}}^2}\left( {{{90}^ \circ } - {{70}^ \circ }} \right)}}$$ $$\eqalign{ & = \frac{{{{\sin }^2}{{25}^ \circ } + {{\cos }^2}{{25}^ \circ }}}{{{{\tan }^2}{{70}^ \circ } - se{c^2}{{70}^ \circ }}} \cr & = \frac{1}{{ - 1}} \cr & = - 1 \cr} $$
67.
If tanÃÂÃÂÃÂø + secÃÂÃÂÃÂø = $$\frac{{x - 2}}{{x + 2}},$$ ÃÂàthen what is the value of cosÃÂÃÂÃÂø?
(A) $$\frac{{{x^2} - 1}}{{{x^2} + 1}}$$
(B) $$\frac{{2{x^2} - 4}}{{2{x^2} + 4}}$$
(C) $$\frac{{{x^2} - 4}}{{{x^2} + 4}}$$
(D) $$\frac{{{x^2} - 2}}{{{x^2} + 2}}$$
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Solution:
$$\eqalign{ & \tan \theta + \sec \theta = \frac{{x - 2}}{{x + 2}}........\left( {\text{i}} \right) \cr & \frac{{\left( {\sec \theta + \tan \theta } \right)}}{1} \times \frac{{\left( {\sec \theta - \tan \theta } \right)}}{{\left( {\sec \theta - \tan \theta } \right)}} = \frac{{x - 2}}{{x + 2}} \cr & \frac{1}{{\sec \theta - \tan \theta }} = \frac{{x - 2}}{{x + 2}} \cr & \left( {\because \,{{\sec }^2}\theta - {{\tan }^2}\theta = 1} \right) \cr & \sec \theta - \tan \theta = \frac{{x + 2}}{{x - 2}}........\left( {{\text{ii}}} \right) \cr & {\text{By equation }}\left( {{\text{ii}}} \right) + \left( {\text{i}} \right){\text{ we get}} \cr & 2\sec \theta = \frac{{x + 2}}{{x - 2}} + \frac{{x - 2}}{{x + 2}} \cr & 2\sec \theta = \frac{{{{\left( {x + 2} \right)}^2} + {{\left( {x - 2} \right)}^2}}}{{\left( {{x^2} - {2^2}} \right)}} \cr & 2\sec \theta = \frac{{\left[ {{x^2} + 4 + 2x + {x^2} + 4 - 2x} \right]}}{{\left( {{x^2} - 4} \right)}} \cr & 2\sec \theta = \frac{{2\left( {{x^2} + 4} \right)}}{{\left( {{x^2} - 4} \right)}} \cr & \sec \theta = \frac{{{x^2} + 4}}{{{x^2} - 4}} \cr & \frac{1}{{\cos \theta }} = \frac{{{x^2} + 4}}{{{x^2} - 4}} \cr & \cos \theta = \frac{{{x^2} - 4}}{{{x^2} + 4}} \cr} $$
68.
The value of sin2 30ÃÂÃÂÃÂð.cos2 45ÃÂÃÂÃÂð + 2tan2 30ÃÂÃÂÃÂð - sec2 60ÃÂÃÂÃÂð is equal to:
(A) $$ - \frac{{13}}{{12}}$$
(B) $$ - \frac{{77}}{{24}}$$
(C) $$ - \frac{{25}}{{12}}$$
(D) $$ - \frac{1}{{12}}$$
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Solution:
$$\eqalign{ & {\sin ^2}{30^ \circ }.{\cos ^2}{45^ \circ } + 2{\tan ^2}{30^ \circ } - {\sec ^2}{60^ \circ } \cr & = {\left( {\frac{1}{2}} \right)^2}.{\left( {\frac{1}{{\sqrt 2 }}} \right)^2} + 2{\left( {\frac{1}{{\sqrt 3 }}} \right)^2} - {\left( 2 \right)^2} \cr & = \frac{1}{4} \times \frac{1}{2} + 2{\left( {\frac{1}{{\sqrt 3 }}} \right)^2} - {\left( 2 \right)^2} \cr & = \frac{1}{8} + \frac{2}{3} - 4 \cr & = \frac{{3 + 16 - 96}}{{24}} \cr & = - \frac{{77}}{{24}} \cr} $$
69.
If 5sinÃÂÃÂÃÂø - 4cosÃÂÃÂÃÂø = 0, 0ÃÂÃÂÃÂð ÃÂÃÂÃÂø 90ÃÂÃÂÃÂð, then the value of $$\frac{{5\sin \theta - 2\cos \theta }}{{5\sin \theta + 3\cos \theta }}$$ ÃÂàis:
(A) $$\frac{3}{7}$$
(B) $$\frac{5}{8}$$
(C) $$\frac{2}{7}$$
(D) $$\frac{3}{8}$$
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Solution:
$$\eqalign{ & 5\sin \theta - 4\cos \theta = 0 \cr & 5\sin \theta = 4\cos \theta \cr & \frac{{\sin \theta }}{{\cos \theta }} = \frac{4}{5} \cr & \tan \theta = \frac{4}{5} \cr & \frac{{5\sin \theta - 2\cos \theta }}{{5\sin \theta + 3\cos \theta }} \cr & = \frac{{5\tan \theta - 2}}{{5\tan \theta + 3}} \cr & = \frac{{5 \times \frac{4}{5} - 2}}{{5 \times \frac{4}{5} + 2}} \cr & = \frac{2}{7} \cr} $$
70.
If tan4 ÃÂÃÂÃÂø + tan2 ÃÂÃÂÃÂø = 1, then the value of cos4 ÃÂÃÂÃÂø + cos2 ÃÂÃÂÃÂø is?
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Solution:
$$\eqalign{ & {\text{ta}}{{\text{n}}^4}\theta + {\text{ta}}{{\text{n}}^2}\theta = 1\,......({\text{i}}) \cr & \because {\sec ^2}\theta - {\text{ta}}{{\text{n}}^2}\theta = 1 \cr & \Rightarrow {\text{ta}}{{\text{n}}^2}\theta \left( {1 + {\text{ta}}{{\text{n}}^2}\theta } \right) = 1 \cr & \Rightarrow {\text{ta}}{{\text{n}}^2}\theta \left( {{{\sec }^2}\theta } \right) = 1 \cr & \Rightarrow {\text{ta}}{{\text{n}}^2}\theta = \frac{1}{{{{\sec }^2}\theta }} \cr & \Rightarrow {\text{ta}}{{\text{n}}^2}\theta = {\text{co}}{{\text{s}}^2}\theta \cr & \because {\text{ co}}{{\text{s}}^4}\theta + {\text{co}}{{\text{s}}^2}\theta \cr & = {\left( {{\text{co}}{{\text{s}}^2}\theta } \right)^2} + {\text{co}}{{\text{s}}^2}\theta \cr & = {\left( {{\text{ta}}{{\text{n}}^2}\theta } \right)^2} + {\text{ta}}{{\text{n}}^2}\theta \cr & = {\text{ta}}{{\text{n}}^4}\theta + {\text{ta}}{{\text{n}}^2}\theta \cr & = 1{\text{ from equation }}\left( {\text{i}} \right) \cr} $$