101.
The value of ÃÂÃÂ ÃÂÃÂ ÃÂÃÂ is?
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Solution:
$$ \Rightarrow {\text{cot1}}{{\text{7}}^ \circ }\left( {\cot {{73}^ \circ }{{\cos }^2}{{22}^ \circ } + \frac{1}{{\cot {{17}^ \circ }se{c^2}{{68}^ \circ }}}} \right)$$ $$ \Rightarrow {\text{cot1}}{{\text{7}}^ \circ }\left[ {\cot \left( {{{90}^ \circ } - {{17}^ \circ }} \right){{\cos }^2}\left( {{{90}^ \circ } - {{68}^ \circ }} \right) + \tan {{17}^ \circ }{\text{co}}{{\text{s}}^2}{{68}^ \circ }} \right]$$ $$\eqalign{ & \Rightarrow {\text{cot1}}{{\text{7}}^ \circ }\left( {tan{{17}^ \circ }si{n^2}{{68}^ \circ } + \tan {{17}^ \circ }{\text{co}}{{\text{s}}^2}{{68}^ \circ }} \right) \cr & \Rightarrow {\text{cot1}}{{\text{7}}^ \circ }\tan {17^ \circ }\left( {si{n^2}{{68}^ \circ } + {\text{co}}{{\text{s}}^2}{{68}^ \circ }} \right) \cr & \Rightarrow 1\left( 1 \right) \cr & \Rightarrow 1 \cr} $$
102.
If ÃÂÃÂ ÃÂÃÂ then find
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Solution:
$$\eqalign{ & \sin \left( {\theta + {{30}^ \circ }} \right) = \frac{3}{{\sqrt {12} }} = \frac{3}{{2\sqrt 3 }} \cr & = \frac{{\sqrt 3 }}{2} = \sin \left( {\theta + {{30}^ \circ }} \right) = \sin {60^ \circ } \cr & \therefore \theta = {30^ \circ } \cr & {\text{co}}{{\text{s}}^2}\theta = {\text{co}}{{\text{s}}^2}{30^ \circ } \cr & = {\left( {\frac{{\sqrt 3 }}{2}} \right)^2} \cr & = \frac{3}{4} \cr} $$
103.
The value of cos2 20ÃÂÃÂÃÂð + cos2 70ÃÂÃÂÃÂð = ?
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Solution:
$$\eqalign{ & = {\cos ^2}{20^ \circ } + {\cos ^2}{70^ \circ } \cr & = {\cos ^2}\left( {{{90}^ \circ } - {{70}^ \circ }} \right) + {\cos ^2}{70^ \circ } \cr & = 1 \cr} $$
104.
If cosÃÂÃÂÃÂø + sinÃÂÃÂÃÂø = cosÃÂÃÂÃÂø, then cosÃÂÃÂÃÂø - sinÃÂÃÂÃÂø is?
(A) tanθ
(B) - cosθ
(C) - sinθ
(D) sinθ
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Solution:
$$\eqalign{ & {\text{cos}}\theta + \sin \theta = \sqrt 2 \cos \theta \cr & {\text{Squaring both sides}} \cr & {\text{co}}{{\text{s}}^2}\theta + {\sin ^2}\theta + 2\cos \theta .\sin \theta = 2{\text{co}}{{\text{s}}^2}\theta \cr & \Rightarrow 2{\text{co}}{{\text{s}}^2}\theta - {\text{co}}{{\text{s}}^2}\theta - {\sin ^2}\theta = 2{\text{cos}}\theta .{\text{sin}}\theta \cr & \Rightarrow {\text{co}}{{\text{s}}^2}\theta - {\sin ^2}\theta = 2\sin \theta .{\text{cos}}\theta \cr & \Rightarrow \left( {\cos \theta - \sin \theta } \right)\left( {\cos \theta + \sin \theta } \right) = 2\sin \theta .{\text{cos}}\theta \cr & \Rightarrow \left( {\cos \theta - \sin \theta } \right)\left( {\sqrt 2 \cos \theta } \right) = 2\sin \theta .{\text{cos}}\theta \cr & \Rightarrow \cos \theta - \sin \theta = \frac{{2\sin \theta .\cos \theta }}{{\sqrt 2 \cos \theta }} \cr & \Rightarrow \sqrt 2 \sin \theta \cr & \cr & {\bf{Alternate:}} \cr & {\text{Let }}\sqrt 2 \cos \theta = \alpha \cr & \therefore \cos \theta \pm \sin \theta = a \cr & \cos \theta \pm \sin \theta = \sqrt {2 - {a^2}} \cr & = \sqrt {2 - {a^2}} \cr & = \sqrt {2 - 2{\text{co}}{{\text{s}}^2}\theta } \cr & = \sqrt {2\left( {1 - {\text{co}}{{\text{s}}^2}\theta } \right)} \cr & = \sqrt {2{{\sin }^2}\theta } \cr & = \sqrt 2 \sin \theta \cr} $$
105.
If ÃÂàÃÂàwhere 0ÃÂÃÂÃÂð ÃÂÃÂÃÂø 90ÃÂÃÂÃÂð, then what is the value of (cos2ÃÂÃÂÃÂø + sin3ÃÂÃÂÃÂø + cosec2ÃÂÃÂÃÂø)?
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Solution:
$$\eqalign{ & \frac{{{{\sin }^2}\theta - 3\sin \theta + 2}}{{{{\cos }^2}\theta }} = 1 \cr & {\text{Put }}\theta = {30^ \circ } \cr & \frac{{{{\sin }^2}{{30}^ \circ } - 3\sin {{30}^ \circ } + 2}}{{{{\cos }^2}{{30}^ \circ }}} = 1 \cr & \frac{1}{4} - \frac{3}{2} + 2 = \frac{3}{4} \cr & \frac{3}{4} = \frac{3}{4} \cr & \cos 2\theta + \sin 3\theta + {\text{cosec}}\,2\theta \cr & = \cos {60^ \circ } + \sin {90^ \circ } + {\text{cosec}}\,{60^ \circ } \cr & = 1 + \frac{1}{2} + \frac{2}{{\sqrt 3 }} \cr & = \frac{{9 + 4\sqrt 3 }}{6} \cr} $$
106.
If tan15ÃÂÃÂÃÂð = 2 - ÃÂàthen the value of tan15ÃÂÃÂÃÂð cot75ÃÂÃÂÃÂð + tan75ÃÂÃÂÃÂð cot15ÃÂÃÂÃÂð is?
(A) 14
(B) 12
(C) 10
(D) 8
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Solution:
$${\text{ tan 1}}{5^ \circ }{\text{cot 7}}{5^ \circ } + {\text{tan 7}}{5^ \circ }{\text{cot 1}}{5^ \circ }$$ $$ = {\text{ tan 1}}{5^ \circ }{\text{cot }}\left( {{{90}^ \circ } - {{15}^ \circ }} \right) + $$ $${\text{tan}}{\left( {{{90}^ \circ } - 15} \right)^ \circ }$$ $${\text{cot1}}{5^ \circ }$$ $$\eqalign{ & = {\text{ ta}}{{\text{n}}^2}{\text{1}}{5^ \circ } + {\text{co}}{{\text{t}}^2}{\text{1}}{5^ \circ } \cr & = {\text{ta}}{{\text{n}}^2}{15^ \circ } + {\text{co}}{{\text{t}}^2}{15^ \circ }\,.....(i) \cr & \left[ {{\bf{Formula}}} \right] \cr & \cot \left( {{{90}^ \circ } - \theta } \right) = \tan \theta \cr & \tan \left( {{{90}^ \circ } - \theta } \right) = \cot \theta \cr & {\text{Put value of tan1}}{5^ \circ } \cr & \cot {15^ \circ } = \frac{1}{{{\text{tan1}}{5^ \circ }}} \cr & \cot {15^ \circ } = \frac{1}{{\left( {2 - \sqrt 3 } \right)}} \cr & \cot {15^ \circ } = \frac{1}{{\left( {2 - \sqrt 3 } \right)}} \times \frac{{\left( {2 + \sqrt 3 } \right)}}{{\left( {2 + \sqrt 3 } \right)}} \cr & \cot {15^ \circ } = 2 + \sqrt 3 \cr & {\text{Now put value in equation (i)}} \cr & {\text{ tan 1}}{5^ \circ } + {\text{cot 1}}{5^ \circ } \cr & = {\left( {2 - \sqrt 3 } \right)^2} + {\left( {2 + \sqrt 3 } \right)^2} \cr & = 4 + 3 - 4\sqrt 3 + 4 + 3 + 4\sqrt 3 \cr & = 14 \cr} $$
107.
If x, y are acute angles, 0 x + y 90ÃÂÃÂÃÂð and sin(2x - 20ÃÂÃÂÃÂð) = cos(2y + 20ÃÂÃÂÃÂð), then the value of tan(x + y) is?
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Solution:
$$\eqalign{ & {\text{sin}}\left( {2x - {{20}^ \circ }} \right) = {\text{cos}}\left( {2y + {{20}^ \circ }} \right) \cr & \Rightarrow \left( {2x - {{20}^ \circ }} \right) + \left( {2y + {{20}^ \circ }} \right) = {90^ \circ } \cr & \left[ {{\text{If sin A}} = {\text{cos B, then A}} + {\text{B}} = {{90}^ \circ }} \right] \cr & \Rightarrow 2\left( {x + y} \right) = {90^ \circ } \cr & \Rightarrow x + y = {45^ \circ } \cr & \therefore \tan \left( {x + y} \right) \cr & = \tan {45^ \circ } \cr & = 1 \cr} $$
108.
If tanÃÂÃÂÃÂø + cotÃÂÃÂÃÂø = 5, then tan2 ÃÂÃÂÃÂø + cot2 ÃÂÃÂÃÂø is?
(A) 23
(B) 24
(C) 25
(D) 26
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Solution:
$$\eqalign{ & {\text{Given,}} \cr & {\text{tan}}\theta + \cot \theta = 5 \cr & \Rightarrow {\text{tan}}\theta + \cot \theta = 5 \cr & \Rightarrow {\left( {{\text{tan}}\theta + \cot \theta } \right)^2} = {5^2} \cr & \left( {{\text{Squaring both sides}}} \right) \cr & \Rightarrow {\text{ta}}{{\text{n}}^2}\theta + {\cot ^2}\theta + 2{\text{tan}}\theta \cot \theta = 25 \cr & \Rightarrow {\text{ta}}{{\text{n}}^2}\theta + {\text{co}}{{\text{t}}^2}\theta = {\text{25}} - {\text{2}} \cr & \left[ {\because {\text{tan}}\theta .\cot \theta = 1} \right] \cr & \Rightarrow {\text{ta}}{{\text{n}}^2}\theta + {\text{co}}{{\text{t}}^2}\theta = 23 \cr} $$
109.
If tanA = n tanB and sinA = m sinB, then the value of cos2 A = ?
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Solution:
$$\eqalign{ & {\text{sin A}} = m{\text{ sin B}} \cr & {\text{si}}{{\text{n}}^2}{\text{A}} = {m^2}{\text{si}}{{\text{n}}^2}{\text{B }}......{\text{(i)}} \cr & {\text{Now, ta}}{{\text{n}}^2}{\text{A}} = {n^2}{\text{ta}}{{\text{n}}^2}{\text{B}} \cr & \frac{{{{\sin }^2}{\text{A}}}}{{{\text{co}}{{\text{s}}^2}{\text{A}}}} = {n^2}\frac{{{{\sin }^2}{\text{B}}}}{{{\text{co}}{{\text{s}}^2}{\text{B}}}} \cr & {\text{from equation (i)}} \cr & \Rightarrow \frac{{1 - {\text{co}}{{\text{s}}^2}{\text{A}}}}{{{{\text{n}}^2}{\text{co}}{{\text{s}}^2}{\text{A}}}} = \frac{{{{\sin }^2}{\text{B}}}}{{{\text{co}}{{\text{s}}^2}{\text{B}}}} \cr & \Rightarrow \frac{{1 - {\text{co}}{{\text{s}}^2}{\text{A}}}}{{{{\text{n}}^2}{\text{co}}{{\text{s}}^2}{\text{A}}}} = \frac{{\frac{{\left( {1 - {\text{co}}{{\text{s}}^2}{\text{A}}} \right)}}{{{m^2}}}}}{{1 - \frac{{{{\sin }^2}{\text{A}}}}{{{m^2}}}}} \cr & \Rightarrow \frac{{1 - {\text{co}}{{\text{s}}^2}{\text{A}}}}{{{{\text{n}}^2}{\text{co}}{{\text{s}}^2}{\text{A}}}} = \frac{{{\text{1}} - {\text{co}}{{\text{s}}^2}{\text{A}}}}{{{m^2} - 1 + {\text{co}}{{\text{s}}^2}{\text{A}}}} \cr & \Rightarrow {m^2} - 1 + {\text{co}}{{\text{s}}^2}{\text{A}} = {{\text{n}}^2}{\text{co}}{{\text{s}}^2}{\text{A}} \cr & \Rightarrow {m^2} - 1 = {\text{co}}{{\text{s}}^2}\theta \left( {{n^2} - 1} \right) \cr & \Rightarrow {\text{co}}{{\text{s}}^2}{\text{A}} = \frac{{{m^2} - 1}}{{{n^2} - 1}} \cr} $$
110.
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:
$$\eqalign{ & {\text{We know that }} \cr & {\text{tan}}\left( {{{90}^ \circ } - \theta } \right) = {\text{cot}}\theta \cr & {\text{and, cot}}\left( {{{90}^ \circ } - \theta } \right) = {\text{tan}}\theta \cr & \Rightarrow {\text{tan}}\left( {4\theta - {{50}^ \circ }} \right) = {\text{cot}}\left( {{{50}^ \circ } - \theta } \right) \cr & \Rightarrow \cot \left[ {{{90}^ \circ } - \left( {4\theta - {{50}^ \circ }} \right)} \right] = {\text{cot}}\left( {{{50}^ \circ } - \theta } \right) \cr & \Rightarrow {90^ \circ } - \left( {4\theta - {{50}^ \circ }} \right) = \left( {{{50}^ \circ } - \theta } \right) \cr & \Rightarrow {90^ \circ } - 4\theta + {50^ \circ } = {50^ \circ } - \theta \cr & \Rightarrow {90^ \circ } = 3\theta \cr & {\text{then}},\theta = {30^ \circ } \cr} $$