51.
If tan ÃÂÃÂÃÂø = 1, then the value of ÃÂàÃÂàis?
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Solution:
$$\eqalign{ & {\text{If }}\tan \theta = 1 \cr & {\text{It means }}\theta = {45^ \circ } \cr & = \frac{{8\sin \theta + 5\cos \theta }}{{{{\sin }^3}\theta - 2{{\cos }^3}\theta + 7\cos \theta }} \cr & = \frac{{8\sin {{45}^ \circ } + 5\cos {{45}^ \circ }}}{{{{\sin }^3}{{45}^ \circ } - 2{{\cos }^3}{{45}^ \circ } + 7\cos {{45}^ \circ }}} \cr & = \frac{{8 \times \frac{1}{{\sqrt 2 }} + 5 \times \frac{1}{{\sqrt 2 }}}}{{{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^3} - 2{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^3} + 7\left( {\frac{1}{{\sqrt 2 }}} \right)}} \cr & = 2 \cr} $$
52.
In a triangle ABC, ÃÂÃÂÃÂâÃÂÃÂÃÂÃÂÃÂàABC = 75ÃÂÃÂÃÂÃÂÃÂÃÂÃÂð and ÃÂÃÂÃÂâÃÂÃÂÃÂÃÂÃÂàACB = ÃÂÃÂÃÂàthe circular measure of ÃÂÃÂÃÂâÃÂÃÂÃÂÃÂÃÂàBAC is?
(A) radian
(B) radian
(C) radian
(D) radian
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Solution:
$$\eqalign{ & \frac{{{\pi ^c}}}{4}{\text{ = }}\frac{{{{180}^ \circ }}}{4}{\text{ = }}{45^ \circ } \cr & \angle {\text{BAC}} = {180^ \circ } - {75^ \circ } - {45^ \circ } = {60^ \circ } \cr & {180^ \circ } \to \pi \cr & {1^ \circ } \to \frac{\pi }{{{{180}^ \circ }}} \cr & {60^ \circ } \to \frac{\pi }{{{{180}^ \circ }}} \times {60^ \circ } = \frac{\pi }{3}{\text{ radian}} \cr} $$
53.
The value of ÃÂÃÂ - ÃÂÃÂ - ÃÂÃÂ is?
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Solution:
$$\eqalign{ & = {\sec ^2}{17^ \circ } - \frac{1}{{{{\tan }^2}{{73}^ \circ }}} - \sin {17^ \circ }\sec {73^ \circ } \cr & = {\sec ^2}{17^ \circ } - {\cot ^2}{73^ \circ } - \sin {17^ \circ }\sec \left( {{{90}^ \circ } - {{17}^ \circ }} \right) \cr & = {\sec ^2}{17^ \circ } - {\cot ^2}\left( {{{90}^ \circ } - {{17}^ \circ }} \right) - \sin {17^ \circ }\cos ec{17^ \circ } \cr & = {\sec ^2}{17^ \circ } - {\tan ^2}{17^ \circ } - 1 \cr & = 1 - 1\left[ {\because {{\sec }^2}\theta - {{\tan }^2}\theta = 1} \right] \cr & = 0 \cr} $$
54.
If ÃÂÃÂ then the value of ÃÂÃÂ is?
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Solution:
$$\eqalign{ & \cot \theta = 4 \cr & \therefore \frac{{5\sin \theta + 3\cos \theta }}{{5\sin \theta - 3\cos \theta }} \cr & = \frac{{5 + 3\cot \theta }}{{5 - 3\cot \theta }} \cr & = \frac{{5 + 3 \times 4}}{{5 - 3 \times 4}} \cr & = - \frac{{17}}{7} \cr} $$
55.
If sec2 A + tan2 A = 3, then what is the value of cotA?
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Solution:
Answer & Solution Answer: Option C No explanation is given for this question Let's Discuss on Board
56.
(A) cosθ(1 + sinθ)
(B) secθ(1 + sinθ)
(C) 2cosθ(1 + secθ)
(D) 2secθ(1 + secθ)
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Solution:
$$\eqalign{ & \frac{{{{\left( {1 + \cos \theta } \right)}^2} + {{\sin }^2}\theta }}{{\left( {{\text{cose}}{{\text{c}}^2}\theta - 1} \right){{\sin }^2}\theta }} \cr & = \frac{{1 + {{\cos }^2}\theta + 2\cos \theta + {{\sin }^2}\theta }}{{\left( {{\text{cose}}{{\text{c}}^2}\theta - 1} \right){{\sin }^2}\theta }} \cr & = \frac{{2\left( {1 + \cos \theta } \right)}}{{\frac{{\left( {1 - {{\sin }^2}\theta } \right)}}{{{{\sin }^2}\theta }}.{{\sin }^2}\theta }} \cr & = \frac{{2\left( {\cos \theta + 1} \right)}}{{{{\cos }^2}\theta }} \cr & = 2\sec \theta \left( {\frac{{\cos \theta }}{{\cos \theta }} + \frac{1}{{\cos \theta }}} \right) \cr & = 2\sec \theta \left( {1 + \sec \theta } \right) \cr} $$
57.
If sin21ÃÂÃÂÃÂð = ÃÂàthen sec21ÃÂÃÂÃÂð - sin69ÃÂÃÂÃÂð is equal to?
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Solution:
$$\eqalign{ & {\text{In }}\vartriangle {\text{ABC sin2}}{{\text{1}}^ \circ }{\text{ = }}\frac{x}{y} \cr & {\text{AB}} = x \cr & {\text{AC}} = y \cr & {\text{BC}} = \sqrt {{y^2} - {x^2}} \cr & \Rightarrow {\text{sec2}}{{\text{1}}^ \circ } - \sin {69^ \circ } \cr & \Rightarrow \frac{{{\text{AC}}}}{{{\text{BC}}}} - \frac{{{\text{BC}}}}{{{\text{AC}}}} \cr & \Rightarrow \frac{{{{\left( {{\text{AC}}} \right)}^2} - {{\left( {{\text{BC}}} \right)}^2}}}{{\left( {{\text{BC}}} \right)\left( {{\text{AC}}} \right)}} = \frac{{{y^2} - {{\left( {\sqrt {\left( {{y^2} - {x^2}} \right)} } \right)}^2}}}{{y\sqrt {{y^2} - {x^2}} }} \cr & \Rightarrow \frac{{{y^2} - {y^2} + {x^2}}}{{y\sqrt {{y^2} + {x^2}} }} = \frac{{{x^2}}}{{y\sqrt {{y^2} - {x^2}} }} \cr} $$
58.
The numerical value of ÃÂÃÂ ÃÂÃÂ is?
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Solution:
$$\eqalign{ & \left( {\frac{1}{{\cos \theta }} + \frac{1}{{\cot \theta }}} \right){\text{ }}\left( {\frac{1}{{\cos \theta }} - \frac{1}{{\cot \theta }}} \right) \cr & = \left( {\sec \theta + \tan \theta } \right)\left( {\sec \theta - \tan \theta } \right) \cr & = {\sec ^2}\theta - {\tan ^2}\theta \left[ {1 + {{\tan }^2}\theta = {{\sec }^2}\theta } \right] \cr & = 1 \cr} $$
59.
The numerical value of 1 + ÃÂÃÂ - ÃÂÃÂ + ÃÂÃÂ - ÃÂÃÂ is?
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Solution:
$$\eqalign{ & {\text{1 + }}\frac{1}{{{\text{co}}{{\text{t}}^2}{{63}^ \circ }}} - {\text{se}}{{\text{c}}^2}{27^ \circ }{\text{ + }}\frac{1}{{{{\sin }^2}{{63}^ \circ }}} - {\text{cose}}{{\text{c}}^2}{27^ \circ } \cr & \Rightarrow 1 + {\text{ta}}{{\text{n}}^2}{63^ \circ } - {\text{se}}{{\text{c}}^2}{27^ \circ } + {\text{cose}}{{\text{c}}^2}{63^ \circ } - {\text{cose}}{{\text{c}}^2}{27^ \circ } \cr & \Rightarrow 1 + {\text{co}}{{\text{t}}^2}{27^ \circ } - {\sec ^2}{27^ \circ } + {\text{se}}{{\text{c}}^2}{27^ \circ } - {\text{cose}}{{\text{c}}^2}{27^ \circ } \cr & \Rightarrow 1 + {\text{co}}{{\text{t}}^2}{27^ \circ } - {\text{cose}}{{\text{c}}^2}{27^ \circ } \cr & \Rightarrow 1 - 1 \cr & \Rightarrow 0 \cr} $$
60.
What is the value of
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Solution:
$$\eqalign{ & \frac{{{{\left[ {\tan \left( {{{90}^ \circ } - A} \right) + \cot \left( {{{90}^ \circ } - A} \right)} \right]}^2}}}{{\left[ {2{{\sec }^2}\left( {{{90}^ \circ } - 2A} \right)} \right]}} \cr & {\text{By putting }}A = {45^ \circ } \cr & \Rightarrow \frac{{{{\left[ {\tan \left( {{{90}^ \circ } - {{45}^ \circ }} \right) + \cot \left( {{{90}^ \circ } - {{45}^ \circ }} \right)} \right]}^2}}}{{\left[ {2{{\sec }^2}\left( {{{90}^ \circ } - 2 \times {{45}^ \circ }} \right)} \right]}} \cr & \Rightarrow \frac{{{{\left[ {\tan {{45}^ \circ } + \cot {{45}^ \circ }} \right]}^2}}}{{2{{\sec }^2}\left( {{{90}^ \circ } - {{90}^ \circ }} \right)}} \cr & \Rightarrow \frac{{{{\left[ {1 + 1} \right]}^2}}}{{2{{\sec }^2}{0^ \circ }}} \cr & \Rightarrow \frac{4}{{2 \times 1}} \cr & \Rightarrow 2 \cr} $$