11.
If sinÃÂÃÂÃÂÃÂÃÂÃÂÃÂø = ÃÂÃÂÃÂâÃÂÃÂÃÂÃÂ3cosÃÂÃÂÃÂÃÂÃÂÃÂÃÂø, 0ÃÂÃÂÃÂÃÂÃÂÃÂÃÂð ÃÂÃÂÃÂÃÂÃÂÃÂÃÂø 90ÃÂÃÂÃÂÃÂÃÂÃÂÃÂð, then the value of 2sinÃÂÃÂÃÂÃÂÃÂÃÂÃÂø + 6secÃÂÃÂÃÂÃÂÃÂÃÂÃÂø + sinÃÂÃÂÃÂÃÂÃÂÃÂÃÂø secÃÂÃÂÃÂÃÂÃÂÃÂÃÂø + cosecÃÂÃÂÃÂÃÂÃÂÃÂÃÂø is22
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Solution:
$$\eqalign{ & \sin \theta = \sqrt 3 \cos \theta \cr & \tan \theta = \sqrt 3 \cr & \tan \theta = \tan {60^ \circ } \cr & \theta = {60^ \circ } \cr & 2{\sin ^2}{60^ \circ } + {\sec ^2}{60^ \circ } + \tan {60^ \circ } + {\text{cosec}}\,{60^ \circ } \cr & = 2 \times \frac{3}{4} + 4 + \sqrt 3 + \frac{2}{{\sqrt 3 }} \cr & = \frac{3}{2} + 4 + \sqrt 3 + \frac{2}{{\sqrt 3 }} \cr & = \frac{{11}}{2} + \frac{5}{{\sqrt 3 }} \cr & = \frac{{11\sqrt 3 + 10}}{{2\sqrt 3 }} \cr & = \frac{{33 + 10\sqrt 3 }}{6} \cr} $$
12.
If ÃÂÃÂÃÂø be an acute angle and 7sin2 ÃÂÃÂÃÂø + 3cos2 ÃÂÃÂÃÂø = 4, then the value of tanÃÂÃÂÃÂø is?
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Solution:
$$\eqalign{ & {\text{7}}{\sin ^2}\theta + 3{\cos ^2}\theta = 4 \cr & \Rightarrow {\text{7}}{\sin ^2}\theta + 3\left( {1 - {{\sin }^2}\theta } \right) = 4 \cr & \Rightarrow {\text{7}}{\sin ^2}\theta + 3 - 3{\sin ^2}\theta = 4 \cr & \Rightarrow 4{\sin ^2}\theta = 1 \cr & \Rightarrow {\sin ^2}\theta = \frac{1}{4} \cr & \Rightarrow \sin \theta = \frac{1}{2} \cr & \Rightarrow \sin \theta = \sin {30^ \circ } \cr & \Rightarrow \theta = {30^ \circ } \cr & \tan {30^ \circ } = \frac{1}{{\sqrt 3 }} \cr & \cr & {\bf{Alternate:}} \cr & {\text{Put}}\theta = {30^ \circ } \cr & {\text{7}} \times \sin^2 {30^ \circ } + 3{\cos ^2}{30^ \circ } = 4 \cr & \Rightarrow 7 \times \frac{1}{4} + 3 \times \frac{3}{4} = 4 \cr & \Rightarrow \frac{7}{4} + \frac{9}{4} = 4 \cr & \Rightarrow \frac{{16}}{4} = 4 \cr & \Rightarrow 4 = 4\left( {{\text{Satisfied}}} \right) \cr & \therefore \tan {30^ \circ } = \frac{1}{{\sqrt 3 }} \cr} $$
13.
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:
$$\eqalign{ & {\text{We know that }} \cr & {\text{tan}}\left( {{{90}^ \circ } - \theta } \right) = {\text{cot}}\theta \cr & {\text{and, cot}}\left( {{{90}^ \circ } - \theta } \right) = {\text{tan}}\theta \cr & \Rightarrow {\text{tan}}\left( {4\theta - {{50}^ \circ }} \right) = {\text{cot}}\left( {{{50}^ \circ } - \theta } \right) \cr & \Rightarrow \cot \left[ {{{90}^ \circ } - \left( {4\theta - {{50}^ \circ }} \right)} \right] = {\text{cot}}\left( {{{50}^ \circ } - \theta } \right) \cr & \Rightarrow {90^ \circ } - \left( {4\theta - {{50}^ \circ }} \right) = \left( {{{50}^ \circ } - \theta } \right) \cr & \Rightarrow {90^ \circ } - 4\theta + {50^ \circ } = {50^ \circ } - \theta \cr & \Rightarrow {90^ \circ } = 3\theta \cr & {\text{then}},\theta = {30^ \circ } \cr} $$
14.
The simplified value of (sec x sec y + tan x tan y)2 - (sec x tan y + tan x sec y)2
(A) -1
(B) 0
(C) sec2 x
(D) 1
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Solution:
(sec x sec y + tan x tan y)2 - (sec x tan y + tan x sec y)2 = sec2x. sec2y + tan2x. tan2y + 2sec x. sec y. tan x. tan y - sec2x. tan2y - tan2x. sec2y + 2sec x. tan y. tan x sec y = sec2x [sec2y - tan2y] - tan2x [sec2y - tan2y] = (sec2x - tan2x) (sec2y - tan2y) = 1 × 1 = 1
15.
The value of 4sin2 30ÃÂÃÂÃÂð + 3cot2 60ÃÂÃÂÃÂð is:
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Solution:
$$\eqalign{ & 4{\sin ^2}{30^ \circ } + 3{\cot ^2}{60^ \circ } \cr & = 4 \times {\left( {\frac{1}{2}} \right)^2} + 3{\left( {\frac{1}{{\sqrt 3 }}} \right)^2} \cr & = 4 \times \frac{1}{4} + 3 \times \frac{1}{3} \cr & = 1 + 1 \cr & = 2 \cr} $$
16.
The value of cot 18ÃÂÃÂÃÂð ÃÂàÃÂàÃÂàis?
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Solution:
$$\cot {18^ \circ }\left( {\cot {{72}^ \circ }.{{\cos }^2}{{22}^ \circ } + \frac{1}{{\tan {{72}^ \circ }.{{\sec }^2}{{68}^ \circ }}}} \right)$$ $$ \Rightarrow \cot {18^ \circ }\left( {\cot {{72}^ \circ }.{{\cos }^2}{{22}^ \circ } + \cot {{72}^ \circ }.{{\cos }^2}{{68}^ \circ }} \right) $$ $$ \Rightarrow \cot {18^ \circ }.\cot {72^ \circ }\left( {{{\cos }^2}{{22}^ \circ } + {{\cos }^2}{{68}^ \circ }} \right) $$ We know that, cotA.cotB = 1 and cos2A + cos2B = 1( when A + B = 90° ) = 1 × 1 = 1
17.
What is the value of
(A) sin cosA
(B) cot
(C) tan
(D) sinAcos
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Solution:
$$\eqalign{ & \frac{{\left[ {\sin \left( {90 - A} \right) + \cos \left( {180 - 2A} \right)} \right]}}{{\left[ {\cos \left( {90 - 2A} \right) + \sin \left( {180 - A} \right)} \right]}} \cr & \Rightarrow \frac{{\cos A + \left( { - \cos 2A} \right)}}{{\sin 2A + \sin A}} \cr & \Rightarrow \frac{{\cos A - \cos 2A}}{{\sin 2A + \sin A}} \cr & \Rightarrow \frac{{ - 2\sin \left( {\frac{{A + 2A}}{2}} \right) \times \sin \left( {\frac{{A - 2A}}{2}} \right)}}{{2\sin \left( {\frac{{A + 2A}}{2}} \right) \times \cos \left( {\frac{{2A - A}}{2}} \right)}} \cr & \Rightarrow \frac{{ - \sin \left( {\frac{{A - 2A}}{2}} \right)}}{{\cos \left( {\frac{{2A - A}}{2}} \right)}} \cr & \Rightarrow \frac{{\sin \left( {\frac{{2A - A}}{2}} \right)}}{{\cos \left( {\frac{{2A - A}}{2}} \right)}} \cr & \Rightarrow \frac{{\sin \left( {\frac{A}{2}} \right)}}{{\cos \left( {\frac{A}{2}} \right)}} \cr & \Rightarrow \tan \left( {\frac{A}{2}} \right) \cr} $$
18.
If ÃÂÃÂ then the value of ÃÂÃÂ is?
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Solution:
$$\eqalign{ & \cot \theta = 4 \cr & \therefore \frac{{5\sin \theta + 3\cos \theta }}{{5\sin \theta - 3\cos \theta }} \cr & = \frac{{5 + 3\cot \theta }}{{5 - 3\cot \theta }} \cr & = \frac{{5 + 3 \times 4}}{{5 - 3 \times 4}} \cr & = - \frac{{17}}{7} \cr} $$
19.
The value of tan 4ÃÂÃÂÃÂð.tan 43ÃÂÃÂÃÂð.tan 47ÃÂÃÂÃÂð.tan 86ÃÂÃÂÃÂð is?
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Solution:
$$\eqalign{ & {\text{tan }}{{\text{4}}^ \circ }{\text{.tan 4}}{{\text{3}}^ \circ }{\text{.tan 4}}{{\text{7}}^ \circ }{\text{.tan 8}}{{\text{6}}^ \circ } \cr & {\text{Here, }} \cr & {\text{tan 8}}{{\text{6}}^ \circ } = {\text{tan}}\left( {{\text{ 9}}{{\text{0}}^ \circ } - {4^ \circ }} \right) = {\text{cot }}{{\text{4}}^ \circ } \cr & {\text{tan 4}}{{\text{7}}^ \circ } = {\text{tan}}\left( {{\text{ 9}}{{\text{0}}^ \circ } - {{43}^ \circ }} \right) = {\text{cot 4}}{{\text{3}}^ \circ } \cr & {\text{tan }}{{\text{4}}^ \circ }{\text{.cot }}{{\text{4}}^ \circ }{\text{.tan 4}}{{\text{3}}^ \circ }{\text{.cot 4}}{{\text{3}}^ \circ } = 1 \cr} $$
20.
If ÃÂÃÂ ÃÂÃÂ = ÃÂÃÂ then m is equal to?
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Solution:
$$\eqalign{ & \frac{{{{\sec }^2}{{70}^ \circ } - {\text{co}}{{\text{t}}^2}{{20}^ \circ }}}{{2\left( {{\text{cose}}{{\text{c}}^2}{{59}^ \circ } - {{\tan }^2}{{31}^ \circ }} \right)}} = \frac{2}{m} \cr & \Rightarrow \frac{{{{\sec }^2}{{70}^ \circ } - {\text{co}}{{\text{t}}^2}\left( {{{90}^ \circ } - {{70}^ \circ }} \right)}}{{2\left( {{\text{cose}}{{\text{c}}^2}{{59}^ \circ } - {{\tan }^2}\left( {{{90}^ \circ } - {{59}^ \circ }} \right)} \right)}} = \frac{2}{m} \cr & \Rightarrow \frac{{{{\sec }^2}{{70}^ \circ } - {\text{ta}}{{\text{n}}^2}{{70}^ \circ }}}{{2\left( {{\text{cose}}{{\text{c}}^2}{{59}^ \circ } - {{\cot }^2}{{59}^ \circ }} \right)}} = \frac{2}{m} \cr & \Rightarrow \frac{1}{2} = \frac{2}{m}\left[ {{{\sec }^2}\theta - {\text{ta}}{{\text{n}}^2}\theta = 1} \right] \cr & (cose{c^2}\theta - {\cot ^2}\theta = 1) \cr & \Rightarrow m = 2 \times 2 \cr & \Rightarrow m = 4 \cr} $$