301.
If $${\text{tan}}\theta = \frac{4}{3}{\text{,}}$$ ÃÂÃÂ then the value of $$\frac{{3\sin \theta + 2{\text{cos}}\theta }}{{3\sin \theta - 2{\text{cos}}\theta }}$$ ÃÂÃÂ is?
(A) 0.5
(B) -0.5
(C) 3.0
(D) -3.0
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Solution:
$$\frac{{3\sin \theta + 2{\text{cos}}\theta }}{{3\sin \theta - 2{\text{cos}}\theta }}$$ Divide numerator & denominator by cosθ $$\eqalign{ & = \frac{{\frac{{3\sin \theta }}{{\cos \theta }} + \frac{{2\cos \theta }}{{\cos \theta }}}}{{\frac{{3\sin \theta }}{{\cos \theta }} - \frac{{2\cos \theta }}{{\cos \theta }}}}\left[ {\frac{{\sin \theta }}{{\cos \theta }} = \tan \theta } \right] \cr & = \frac{{3\tan \theta + 2}}{{3\tan \theta - 2}} \cr & {\text{Put value of tan}}\theta \cr & = \frac{{3 \times \frac{4}{3} + 2}}{{3 \times \frac{4}{3} - 2}} \cr & = \frac{6}{2} \cr & = 3 \cr} $$
302.
If cos(A - B) = $$\frac{{\sqrt 3 }}{2}$$ and sec A = 2, 0ÃÂÃÂÃÂÃÂÃÂÃÂÃÂð ÃÂÃÂÃÂâÃÂÃÂÃÂÃÂÃÂä A ÃÂÃÂÃÂâÃÂÃÂÃÂÃÂÃÂä 90ÃÂÃÂÃÂÃÂÃÂÃÂÃÂð, 0ÃÂÃÂÃÂÃÂÃÂÃÂÃÂð ÃÂÃÂÃÂâÃÂÃÂÃÂÃÂÃÂä B ÃÂÃÂÃÂâÃÂÃÂÃÂÃÂÃÂä 90ÃÂÃÂÃÂÃÂÃÂÃÂÃÂð then what is the measure of B?
(A) 60°
(B) 0°
(C) 30°
(D) 90°
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Solution:
$$\eqalign{ & \cos \left( {A - B} \right) = \frac{{\sqrt 3 }}{2} \cr & \cos \left( {A - B} \right) = \cos {30^ \circ } \cr & A - B = {30^ \circ }........\left( {\text{i}} \right) \cr & \sec A = 2 \cr & \cos A = \frac{1}{2} = \cos {60^ \circ } \cr & A = {60^ \circ }........\left( {{\text{ii}}} \right) \cr & {\text{From equation }}\left( {\text{i}} \right){\text{ and }}\left( {{\text{ii}}} \right) \cr & A = {60^ \circ }\,\& \,B = {30^ \circ } \cr} $$
303.
The value of $$\frac{{\sec \theta \left( {1 - \sin \theta } \right)\left( {\sin \theta + \cos \theta } \right)\left( {\sec \theta + \tan \theta } \right)}}{{\sin \theta \left( {1 + \tan \theta } \right) + \cos \theta \left( {1 + \cot \theta } \right)}}$$ ÃÂÃÂ ÃÂÃÂ ÃÂÃÂ ÃÂÃÂ is equal to:
(A) 2cosθ
(B) cosecθsecθ
(C) 2sinθ
(D) sinθcosθ
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Solution:
$$\eqalign{ & \frac{{\sec \theta \left( {1 - \sin \theta } \right)\left( {\sin \theta + \cos \theta } \right)\left( {\sec \theta + \tan \theta } \right)}}{{\sin \theta \left( {1 + \tan \theta } \right) + \cos \theta \left( {1 + \cot \theta } \right)}} \cr & = \frac{{\left( {\sec \theta - \tan \theta } \right)\left( {\sin \theta + \cos \theta } \right)\left( {\sec \theta + \tan \theta } \right)}}{{\sin \theta \left( {1 + \tan \theta } \right) + \cos \theta \left( {\frac{{1 + \tan \theta }}{{\tan \theta }}} \right)}} \cr & = \frac{{\left( {{{\sec }^2}\theta - {{\tan }^2}\theta } \right)\left( {\sin \theta + \cos \theta } \right)}}{{\left( {1 + \tan \theta } \right)\left( {\sin \theta + \frac{{{{\cos }^2}\theta }}{{\sin \theta }}} \right)}} \cr & = \frac{{1\left( {\sin \theta + \cos \theta } \right)}}{{\left( {1 + \tan \theta } \right)\left( {\frac{1}{{\sin \theta }}} \right)}} \cr & = \frac{{\left( {\sin \theta + \cos \theta } \right)}}{{\left( {\cos \theta + \sin \theta } \right)}}\sin \theta .\cos \theta \cr & = \sin \theta .\cos \theta \cr} $$
304.
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?
(A) $$\frac{1}{{\sqrt 3 }}$$
(B) $$\frac{{\sqrt 3 }}{2}$$
(C) $$\sqrt 3 $$
(D) 1
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Solution:
$$\eqalign{ & {\text{sin}}\left( {2x - {{20}^ \circ }} \right) = {\text{cos}}\left( {2y + {{20}^ \circ }} \right) \cr & \Rightarrow \left( {2x - {{20}^ \circ }} \right) + \left( {2y + {{20}^ \circ }} \right) = {90^ \circ } \cr & \left[ {{\text{If sin A}} = {\text{cos B, then A}} + {\text{B}} = {{90}^ \circ }} \right] \cr & \Rightarrow 2\left( {x + y} \right) = {90^ \circ } \cr & \Rightarrow x + y = {45^ \circ } \cr & \therefore \tan \left( {x + y} \right) \cr & = \tan {45^ \circ } \cr & = 1 \cr} $$
305.
If ÃÂÃÂÃÂñ and ÃÂÃÂÃÂò are positive acute angles, sin(4ÃÂÃÂÃÂñ - ÃÂÃÂÃÂò) = 1 and cos(2ÃÂÃÂÃÂñ + ÃÂÃÂÃÂò) = $$\frac{1}{2}{\text{,}}$$ then the value of sin(ÃÂÃÂÃÂñ + 2ÃÂÃÂÃÂò) is?
(A) 0
(B) 1
(C) $$\frac{{\sqrt 3 }}{2}$$
(D) $$\frac{1}{{\sqrt 2 }}$$
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Solution:
$$\eqalign{ & {\text{sin}}\left( {4\alpha - \beta } \right) = 1 = \sin {90^ \circ } \cr & 4\alpha - \beta = {90^ \circ }{\text{ }} \cr & {\text{cos}}\left( {2\alpha + \beta } \right) = \frac{1}{2} = \cos {60^ \circ } \cr & 2\alpha + \beta = {60^ \circ } \cr & {\text{Adding }}6\alpha = {150^ \circ } \cr & \alpha = {25^ \circ } \cr & \beta = {10^ \circ } \cr & \Rightarrow {\text{sin}}\left( {\alpha + 2\beta } \right) \cr & \Rightarrow {\text{sin}}\left( {{{25}^ \circ } + 2 \times {{10}^ \circ }} \right) \cr & \Rightarrow {\text{sin}}{45^ \circ } \cr & \Rightarrow \frac{1}{{\sqrt 2 }} \cr} $$
306.
If $$\theta $$ is a positive acute angle and $${\text{4}}{\sin ^2}\theta $$ ÃÂÃÂ = 3, then the value of $${\text{tan}}\theta $$ÃÂÃÂ - $$cot\frac{\theta }{2}$$ ÃÂÃÂ is?
(A) 1
(B) 0
(C) $$\sqrt 3 $$
(D) $$\frac{1}{{\sqrt 3 }}$$
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Solution:
$$\eqalign{ & {\text{Given 4}}{\sin ^2}\theta = 3 \cr & {\sin ^2}\theta = \frac{3}{4} \cr & \sin \theta = \frac{{\sqrt 3 }}{2} \cr & \sin \theta = {\text{sin }}{60^ \circ } \cr & \theta = {60^ \circ } \cr & \because \tan \theta - \cot \frac{\theta }{2} \cr & = \tan {60^ \circ } - \cot \frac{{{{60}^ \circ }}}{2} \cr & = \tan {60^ \circ } - \cot {30^ \circ } \cr & = \sqrt 3 - \sqrt 3 \cr & = 0 \cr} $$
307.
If ÃÂÃÂÃÂø is positive acute angle and 7cos2 ÃÂÃÂÃÂø + 3sin2 ÃÂÃÂÃÂø = 4, then value of ÃÂÃÂÃÂø is?
(A) 60°
(B) 30°
(C) 45°
(D) 90°
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Solution:
$$\eqalign{ & {\text{7co}}{{\text{s}}^2}\theta + 3{\sin ^2}\theta = 4 \cr & 7{\text{co}}{{\text{s}}^2}\theta + 3\left( {1 - {\text{co}}{{\text{s}}^2}\theta } \right) - 4 = 0 \cr & 7{\text{co}}{{\text{s}}^2}\theta - 3{\text{co}}{{\text{s}}^2}\theta + 3 - 4 = 0 \cr & 4{\text{co}}{{\text{s}}^2}\theta = 1 \cr & {\text{co}}{{\text{s}}^2}\theta = \frac{1}{4} \cr & {\text{cos}}\theta = \frac{1}{2} \cr & \cos \theta = {\text{cos}}{60^ \circ } \cr & \theta = {60^ \circ } \cr} $$
308.
The value of cotÃÂÃÂÃÂø.tan(90ÃÂÃÂÃÂð - ÃÂÃÂÃÂø) - sec(90ÃÂÃÂÃÂð - ÃÂÃÂÃÂø)cosecÃÂÃÂÃÂø + (sin2 25ÃÂÃÂÃÂð + sin2 65ÃÂÃÂÃÂð) + $$\sqrt 3 $$ (tan5ÃÂÃÂÃÂð. tan15ÃÂÃÂÃÂð. tan30ÃÂÃÂÃÂð. tan75ÃÂÃÂÃÂð. tan85ÃÂÃÂÃÂð)
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Solution:
cotθ. tan(90° - θ) - sec(90° - θ)cosecθ + (sin225° + sin265°) + $$\sqrt 3 $$ (tan5°. tan15°. tan30°. tan75°. tan85°) = cotθ. cotθ - cosecθ. cosecθ + (sin225° + cos225° ) + $$\sqrt 3 $$ [(tan5°. tan85°) . (tan15°. tan75°). tan30°] $$ = \left( {{\text{co}}{{\text{t}}^2}\theta - {\text{cose}}{{\text{c}}^2}\theta } \right) + \left( 1 \right) + \sqrt 3 \left( {1.1.\frac{1}{{\sqrt 3 }}} \right)$$ $$\left[ {{\text{tan A}}{\text{.tan B}} = {\text{1, If A}} + {\text{B}} = {{90}^ \circ }} \right]$$ $$\eqalign{ & = \left( { - 1} \right) + \left( 1 \right) + \sqrt 3 \times \frac{1}{{\sqrt 3 }} \cr & = - 1 + 1 + 1 \cr & = 1 \cr} $$
309.
0 ÃÂÃÂÃÂÃÂÃÂÃÂÃÂø 90ÃÂÃÂÃÂÃÂÃÂÃÂÃÂð, tanÃÂÃÂÃÂÃÂÃÂÃÂÃÂø + sinÃÂÃÂÃÂÃÂÃÂÃÂÃÂø =m and tanÃÂÃÂÃÂÃÂÃÂÃÂÃÂø - sinÃÂÃÂÃÂÃÂÃÂÃÂÃÂø = n, where m ÃÂÃÂÃÂâÃÂÃÂÃÂÃÂÃÂàn, then the value of m - n is?22
(A) 2(tan2θ + sin2θ)
(B) 4mn
(C) 4$$\sqrt {\text{mn}} $$
(D) 2(m2 + n2)
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Solution:
$$\eqalign{ & \tan \theta + \sin \theta = m \cr & {\text{Squaring both sides}} \cr & {\tan ^2}\theta + {\sin ^2}\theta + 2{\text{ tan}}\theta .\sin \theta = {m^2}\,....(i) \cr & {\text{tan}}\theta - \sin \theta = n \cr & {\text{Squaring both sides}} \cr & {\tan ^2}\theta + {\sin ^2}\theta - 2{\text{ tan}}\theta .\sin \theta = {n^2}\,....(ii) \cr & {\text{Substract from (i) and (ii)}} \cr & {m^2} - {n^2} = {\text{ta}}{{\text{n}}^2}\theta + {\sin ^2}\theta + 2{\text{tan}}\theta \sin \theta - {\text{ta}}{{\text{n}}^2}\theta - {\sin ^2}\theta + 2{\text{tan}}\theta \sin \theta \cr & {m^2} - {n^2} = 4{\text{tan}}\theta \sin \theta \cr & = 4\sqrt {{\text{ta}}{{\text{n}}^2}\theta {{\sin }^2}\theta } \cr & = 4\sqrt {{\text{ta}}{{\text{n}}^2}\theta \left( {1 - {\text{co}}{{\text{s}}^2}\theta } \right)} \cr & = 4\sqrt {{\text{ta}}{{\text{n}}^2}\theta - {{\sin }^2}\theta } \cr & = 4\sqrt {mn} \cr} $$
310.
The value of $$\frac{{4{{\tan }^2}{{30}^ \circ } + {{\sin }^2}{{30}^ \circ }{{\cos }^2}{{45}^ \circ } + {{\sec }^2}{{48}^ \circ } - {{\cot }^2}{{42}^ \circ }}}{{\cos {{37}^ \circ }\sin {{53}^ \circ } + \sin {{37}^ \circ }\cos {{53}^ \circ } + \tan {{18}^ \circ }\tan {{72}^ \circ }}}\,{\text{is:}}$$
(A) $$\frac{{49}}{{24}}$$
(B) $$\frac{{35}}{{48}}$$
(C) $$\frac{{35}}{{24}}$$
(D) $$\frac{{59}}{{48}}$$
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Solution:
$$\eqalign{ & \frac{{4{{\tan }^2}{{30}^ \circ } + {{\sin }^2}{{30}^ \circ }{{\cos }^2}{{45}^ \circ } + {{\sec }^2}{{48}^ \circ } - {{\cot }^2}{{42}^ \circ }}}{{\cos {{37}^ \circ }\sin {{53}^ \circ } + \sin {{37}^ \circ }\cos {{53}^ \circ } + \tan {{18}^ \circ }\tan {{72}^ \circ }}} \cr & = \frac{{4{{\left( {\frac{1}{{\sqrt 3 }}} \right)}^2} + {{\left( {\frac{1}{2}} \right)}^2}{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2} + {\text{cose}}{{\text{c}}^2}{{42}^ \circ } - {{\cot }^2}{{42}^ \circ }}}{{\cos {{37}^ \circ }\cos {{37}^ \circ } + \sin {{37}^ \circ }\sin {{37}^ \circ } + 1}} \cr & = \frac{{4 \times \frac{1}{3} + \frac{1}{4} \times \frac{1}{2} + 1}}{{{{\cos }^2}{{37}^ \circ } + {{\sin }^2}{{37}^ \circ } + 1}} \cr & = \frac{{\frac{4}{3} + \frac{1}{8} + 1}}{{1 + 1}} \cr & = \frac{{32 + 3 + 24}}{{24 \times 2}} \cr & = \frac{{59}}{{48}} \cr} $$