131.
If xsin60ÃÂÃÂÃÂð.tan30ÃÂÃÂÃÂð - tan2 45ÃÂÃÂÃÂð = cosec60ÃÂÃÂÃÂð.cot30ÃÂÃÂÃÂð - sec2 45ÃÂÃÂÃÂð then x = ?
(A) 2
(B) -2
(C) 6
(D) -4
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Solution:
$${\text{ }}x\sin {60^ \circ }.\tan {30^ \circ } - \tan^2 {45^ \circ } = $$ $$\operatorname{cosec} {60^ \circ }.$$ $$\cot {30^ \circ } - $$ $${\sec ^2}{45^ \circ }$$ $$\eqalign{ & \Rightarrow x \times \frac{{\sqrt 3 }}{2} \times \frac{1}{{\sqrt 3 }} - 1 = \frac{2}{{\sqrt 3 }} \times \sqrt 3 - {\left( {\sqrt 2 } \right)^2} \cr & \Rightarrow \frac{x}{2} - 1 = 2 - 2 \cr & \Rightarrow \frac{x}{2} - 1 = 0 \cr & \Rightarrow \frac{x}{2} = 1 \cr & \Rightarrow x = 2 \cr} $$
132.
The value of sin2 30ÃÂÃÂÃÂð.cos2 45ÃÂÃÂÃÂð + 2tan2 30ÃÂÃÂÃÂð - sec2 60ÃÂÃÂÃÂð is equal to:
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Solution:
$$\eqalign{ & {\sin ^2}{30^ \circ }.{\cos ^2}{45^ \circ } + 2{\tan ^2}{30^ \circ } - {\sec ^2}{60^ \circ } \cr & = {\left( {\frac{1}{2}} \right)^2}.{\left( {\frac{1}{{\sqrt 2 }}} \right)^2} + 2{\left( {\frac{1}{{\sqrt 3 }}} \right)^2} - {\left( 2 \right)^2} \cr & = \frac{1}{4} \times \frac{1}{2} + 2{\left( {\frac{1}{{\sqrt 3 }}} \right)^2} - {\left( 2 \right)^2} \cr & = \frac{1}{8} + \frac{2}{3} - 4 \cr & = \frac{{3 + 16 - 96}}{{24}} \cr & = - \frac{{77}}{{24}} \cr} $$
133.
The value of
(A) secθ
(B) sinθ
(C) cotθ
(D) tanθ
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Solution:
$$\eqalign{ & \frac{{2{{\cos }^3}\theta - \cos \theta }}{{\sin \theta - 2{{\sin }^3}\theta }} \cr & = \frac{{\cos \theta \left[ {2{{\cos }^2}\theta - 1} \right]}}{{\sin \theta \left[ {1 - 2{{\sin }^2}\theta } \right]}} \cr & = \frac{{\cos \theta \times \cos 2\theta }}{{\sin \theta \times \cos 2\theta }} \cr & = \cot \theta \cr} $$
134.
If cosÃÂÃÂÃÂÃÂÃÂÃÂÃÂø - sinÃÂÃÂÃÂÃÂÃÂÃÂÃÂø = where 0 ÃÂÃÂÃÂâÃÂÃÂÃÂÃÂÃÂä ÃÂÃÂÃÂÃÂÃÂÃÂÃÂø ÃÂÃÂÃÂâÃÂÃÂÃÂÃÂÃÂä then the value of cosÃÂÃÂÃÂÃÂÃÂÃÂÃÂø - sinÃÂÃÂÃÂÃÂÃÂÃÂÃÂø is?2244
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Solution:
$$\eqalign{ & {\text{co}}{{\text{s}}^2}\theta - {\sin ^2}\theta = \frac{1}{3}{\text{ }}\left( {{\text{Given}}} \right) \cr & {\text{co}}{{\text{s}}^2}\theta + {\sin ^2}\theta = 1{\text{ }}\left( {{\text{Property}}} \right) \cr & \left( {{\text{co}}{{\text{s}}^2}\theta - {{\sin }^2}\theta } \right)\left( {{\text{co}}{{\text{s}}^2}\theta + {{\sin }^2}\theta } \right) = \frac{1}{3} \times 1 \cr & {\text{co}}{{\text{s}}^4}\theta - {\sin ^4}\theta = \frac{1}{3} \cr & \therefore \left( {\left( {{a^2} + {b^2}} \right)\left( {{a^2} - {b^2}} \right) = {a^4} - {b^4}} \right) \cr} $$
135.
If cos53ÃÂÃÂÃÂð = then sec53ÃÂÃÂÃÂð + cot37ÃÂÃÂÃÂð is equal to:
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Solution:
$$\eqalign{ & \cot {53^ \circ } = \frac{x}{y} = \frac{B}{H} \cr & {P^2} = {H^2} - {B^2} \cr & P = \sqrt {{y^2} - {x^2}} \cr & \sec {53^ \circ } + \cot {37^ \circ } \cr & = \sec {53^ \circ } + \tan {53^ \circ } \cr & = \frac{H}{B} + \frac{P}{B} \cr & = \frac{{H + P}}{B} \cr & = \frac{{y + \sqrt {{y^2} - {x^2}} }}{x} \cr} $$
136.
If cosÃÂÃÂÃÂø.cosec23ÃÂÃÂÃÂð = 1, the value of ÃÂÃÂÃÂø is?
(A) 23°
(B) 37°
(C) 63°
(D) 67°
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Solution:
$$\eqalign{ & {\text{cos}}\theta .{\text{cosec2}}{{\text{3}}^ \circ } = {\text{1}} \cr & {\text{If cosA}}.{\text{cosecB}} = {\text{1}} \cr & {\text{Then, A + B}} = {90^ \circ } \cr & {\text{So,}} \cr & \theta {\text{ + 2}}{{\text{3}}^ \circ } = {90^ \circ } \cr & \theta = {67^ \circ } \cr} $$
137.
If sinÃÂÃÂÃÂø = 4cosÃÂÃÂÃÂø, then what is the value of sinÃÂÃÂÃÂøcosÃÂÃÂÃÂø?
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Solution:
$$\eqalign{ & \sin \theta = 4\cos \theta \cr & \tan \theta = 4 \cr & \sin \theta .\cos \theta = \frac{4}{{\sqrt {17} }} \times \frac{1}{{\sqrt {17} }} = \frac{4}{{17}} \cr} $$
138.
The expression (tanÃÂÃÂÃÂø + cotÃÂÃÂÃÂø)(secÃÂÃÂÃÂø + tanÃÂÃÂÃÂø)(1 - sinÃÂÃÂÃÂø), 0ÃÂÃÂÃÂð ÃÂÃÂÃÂø 90ÃÂÃÂÃÂð is equal to:
(A) sinθ
(B) secθ
(C) cosecθ
(D) cotθ
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Solution:
$$\eqalign{ & \left( {\tan \theta + \cot \theta } \right)\left( {\sec \theta + \tan \theta } \right)(1 - \sin \theta ) \cr & = \left( {\frac{{\sin \theta }}{{\cos \theta }} + \frac{{\cos \theta }}{{\sin \theta }}} \right)\left( {\frac{1}{{\cos \theta }} + \frac{{\sin \theta }}{{\cos \theta }}} \right)\left( {1 - \sin \theta } \right) \cr & = \left( {\frac{{{{\sin }^2}\theta + {{\cos }^2}\theta }}{{\cos \theta \sin \theta }}} \right)\left( {\frac{{1 + \sin \theta }}{{\cos \theta }}} \right)\left( {1 - \sin \theta } \right) \cr & = \left( {\frac{1}{{\cos \theta \sin \theta }}} \right)\left( {\frac{{1 - {{\sin }^2}\theta }}{{\cos \theta }}} \right) \cr & = \left( {\frac{1}{{\cos \theta \sin \theta }}} \right)\left( {\frac{{{{\cos }^2}\theta }}{{\cos \theta }}} \right) \cr & = \frac{1}{{\sin \theta }} \cr & = {\text{cosec}}\,\theta \cr} $$
139.
The value of the expression 2(sin6 ÃÂÃÂÃÂø + cos6 ÃÂÃÂÃÂø) - 3(sin4 ÃÂÃÂÃÂø + cos4 ÃÂÃÂÃÂø) + 1 is?
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Solution:
$${\text{ 2}}\left( {{{\sin }^6}\theta + {\text{co}}{{\text{s}}^6}\theta } \right) - {\text{3}}\left( {{{\sin }^4}\theta + {\text{co}}{{\text{s}}^4}\theta } \right){\text{ + 1}}$$ $$ \Rightarrow {\text{2}}\left( {1 - 3{{\sin }^2}\theta {\text{co}}{{\text{s}}^2}\theta } \right) - $$ $${\text{3}}\left( {1 - 2{{\sin }^2}\theta c{\text{o}}{{\text{s}}^2}\theta } \right){\text{ + }}$$ $${\text{1}}$$ $$\eqalign{ & \Rightarrow {\text{ 2}} - 6{\sin ^2}\theta .{\text{co}}{{\text{s}}^2}\theta - 3 + 6{\sin ^2}\theta .{\text{co}}{{\text{s}}^2}\theta {\text{ + 1}} \cr & \Rightarrow 2 - 3 + 1 \cr & \Rightarrow 0 \cr} $$
140.
If ÃÂÃÂÃÂâÃÂÃÂÃÂÃÂÃÂàA and ÃÂÃÂÃÂâÃÂÃÂÃÂÃÂÃÂàB are complementary to each other, then the value of secA + secB - secA.secB is?2222
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Solution:
$$\eqalign{ & {\text{A + B}} = {90^ \circ } \cr & {\text{B}} = {90^ \circ } - {\text{A}} \cr & {\sec ^2}A + se{c^2}{\text{B}} - {\text{se}}{{\text{c}}^2}{\text{A}}{\text{.se}}{{\text{c}}^2}{\text{B}} \cr} $$ $$ = {\sec ^2}A + se{c^2}\left( {{{90}^ \circ } - {\text{A}}} \right) - $$ $${\text{se}}{{\text{c}}^2}{\text{A}}{\text{.se}}{{\text{c}}^2}$$ $$\left( {{{90}^ \circ } - {\text{A}}} \right)$$ $$\eqalign{ & = {\sec ^2}A + {\operatorname{cosec} ^2}{\text{A}} - {\text{se}}{{\text{c}}^2}{\text{A}}{\text{.cose}}{{\text{c}}^2}{\text{A}} \cr & = \frac{1}{{{{\cos }^2}{\text{A}}}} + \frac{1}{{si{n^2}{\text{A}}}} - \frac{1}{{{\text{co}}{{\text{s}}^2}{\text{A}}}} \times \frac{1}{{{\text{si}}{{\text{n}}^2}{\text{A}}}} \cr & = \frac{{{{\sin }^2}{\text{A + co}}{{\text{s}}^2}{\text{A}}}}{{{\text{co}}{{\text{s}}^2}{\text{A}}{\text{.}}{{\sin }^2}{\text{A}}}} - \frac{1}{{{\text{co}}{{\text{s}}^2}{\text{A}}{\text{.}}{{\sin }^2}{\text{A}}}} \cr & = \frac{1}{{{\text{co}}{{\text{s}}^2}{\text{A}}.{{\sin }^2}{\text{A}}}} - \frac{1}{{{\text{co}}{{\text{s}}^2}{\text{A}}.{{\sin }^2}{\text{A}}}} \cr & = 0 \cr} $$