81.
The value of ÃÂàÃÂàif 0ÃÂÃÂÃÂð ÃÂÃÂÃÂø 90ÃÂÃÂÃÂð, is equal to:
(A) 1 - secθ + tanθ
(B) 1 - secθ - tanθ
(C) 1 + secθ - tanθ
(D) 1 + secθ + tanθ
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Solution:
$$\eqalign{ & 1 + \sqrt {\frac{{\cot \theta + \cos \theta }}{{\cot \theta - \cos \theta }}} \cr & = 1 + \sqrt {\frac{{\frac{{\cos \theta }}{{\sin \theta }} + \cos \theta }}{{\frac{{\cos \theta }}{{\sin \theta }} - \cos \theta }}} \cr & = 1 + \sqrt {\frac{{\cot \theta + \sin \theta \cos \theta }}{{\cot \theta - \cos \theta \sin \theta }}} \cr & = 1 + \sqrt {\frac{{\cos \theta \left( {1 + \sin \theta } \right)}}{{\cos \theta \left( {1 - \sin \theta } \right)}}} \cr & = 1 + \sqrt {\frac{{1 + \sin \theta }}{{1 - \sin \theta }}} \cr & {\text{Rationalise,}} \cr & = 1 + \sqrt {\frac{{1 + \sin \theta }}{{1 - \sin \theta }} \times \frac{{1 + \sin \theta }}{{1 + \sin \theta }}} \cr & = 1 + \sqrt {\frac{{{{\left( {1 + \sin \theta } \right)}^2}}}{{1 - {{\sin }^2}\theta }}} \cr & = 1 + \sqrt {{{\left( {1 + \sin \theta } \right)}^2} \times {{\sec }^2}\theta } \cr & = 1 + \left( {1 + \sin \theta } \right)\sec \theta \cr & = \sec \theta + \sin \theta \sec \theta \cr & = 1 + \sec \theta + \tan \theta \cr} $$
82.
If tanÃÂÃÂÃÂø = tan30ÃÂÃÂÃÂð .tan60ÃÂÃÂÃÂð and ÃÂÃÂÃÂø is an acute angle, then 2ÃÂÃÂÃÂø is equal to?
(A) 30°
(B) 45°
(C) 90°
(D) 0°
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Solution:
$$\eqalign{ & {\text{tan}}\theta = {\text{tan}}{30^ \circ }.{\text{tan}}{60^ \circ } \cr & {\text{tan}}\theta = \frac{1}{{\sqrt 3 }}.\sqrt 3 \cr & {\text{tan}}\theta = 1 \cr & {\text{tan}}\theta = {\text{tan}}{45^ \circ } \cr & \theta = {45^ \circ } \cr & \therefore 2\theta = {90^ \circ } \cr} $$
83.
a, b, c are the lengths of three sides of a triangle ABC. If a, b, c are related by the relation a2 + b2 + c2 = ab + bc + ca, then the value of (sin2 A + sin2 B + sin2 C) is?
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Solution:
$$\eqalign{ & {a^2} + {b^2} + {c^2} = ab + bc + ca \cr & \Rightarrow {a^2} + {b^2} + {c^2} - ab - bc - ca = 0 \cr & \Rightarrow 2{a^2} + 2{b^2} + 2{c^2} - 2ab - 2bc - 2ca = 0 \cr & \Rightarrow {a^2} + {b^2} - 2ab + {b^2} + {c^2} - 2bc + {c^2} + {a^2} - 2ca = 0 \cr & \Rightarrow {\left( {a - b} \right)^2} + {\left( {b - c} \right)^2} + {\left( {c - a} \right)^2} = 0 \cr & \therefore a = b = c \cr & \vartriangle {\text{ABC}} = {\text{equilateral }}\vartriangle \cr & \therefore \angle {\text{A}} = \angle {\text{B}} = \angle {\text{C}} = {60^ \circ } \cr & {\text{So, }}{\sin ^2}{\text{A}} + {\sin ^2}{\text{B}} + {\sin ^2}{\text{C}} \cr & \Rightarrow {\sin ^2}{60^ \circ } + {\sin ^2}{60^ \circ } + {\sin ^2}{60^ \circ } \cr & \Rightarrow 3{\sin ^2}{60^ \circ } \cr & \Rightarrow 3 \times {\left( {\frac{{\sqrt 3 }}{2}} \right)^2} \cr & \Rightarrow \frac{9}{4} \cr} $$
84.
If ÃÂÃÂ ÃÂÃÂ = ÃÂÃÂ then m is equal to?
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Solution:
$$\eqalign{ & \frac{{{{\sec }^2}{{70}^ \circ } - {\text{co}}{{\text{t}}^2}{{20}^ \circ }}}{{2\left( {{\text{cose}}{{\text{c}}^2}{{59}^ \circ } - {{\tan }^2}{{31}^ \circ }} \right)}} = \frac{2}{m} \cr & \Rightarrow \frac{{{{\sec }^2}{{70}^ \circ } - {\text{co}}{{\text{t}}^2}\left( {{{90}^ \circ } - {{70}^ \circ }} \right)}}{{2\left( {{\text{cose}}{{\text{c}}^2}{{59}^ \circ } - {{\tan }^2}\left( {{{90}^ \circ } - {{59}^ \circ }} \right)} \right)}} = \frac{2}{m} \cr & \Rightarrow \frac{{{{\sec }^2}{{70}^ \circ } - {\text{ta}}{{\text{n}}^2}{{70}^ \circ }}}{{2\left( {{\text{cose}}{{\text{c}}^2}{{59}^ \circ } - {{\cot }^2}{{59}^ \circ }} \right)}} = \frac{2}{m} \cr & \Rightarrow \frac{1}{2} = \frac{2}{m}\left[ {{{\sec }^2}\theta - {\text{ta}}{{\text{n}}^2}\theta = 1} \right] \cr & (cose{c^2}\theta - {\cot ^2}\theta = 1) \cr & \Rightarrow m = 2 \times 2 \cr & \Rightarrow m = 4 \cr} $$
85.
The value of
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Solution:
$$\eqalign{ & \left( {\frac{{\sin A}}{{1 - \cos A}} + \frac{{1 - \cos A}}{{\sin A}}} \right) \div \left( {\frac{{{{\cot }^2}A}}{{1 + {\text{cosec}}\,A}} + 1} \right) \cr & = \left( {\frac{{{{\sin }^2}A + {{\left( {1 + \cos A} \right)}^2}}}{{\sin A\left( {1 - \cos A} \right)}}} \right) \div \left( {\frac{{\frac{{{{\cos }^2}A}}{{{{\sin }^2}A}} \times \sin A}}{{1 + \sin A}} + 1} \right) \cr & = \frac{{{{\sin }^2}A + 1 + {{\cos }^2}A - 2\cos A}}{{\sin A\left( {1 - \cos A} \right)}} \div \frac{{{{\cos }^2}A + \sin A + {{\sin }^2}A}}{{\sin A\left( {1 + \sin A} \right)}} \cr & = \frac{{2\left( {1 - \cos A} \right)}}{{\sin A\left( {1 - \cos A} \right)}} \times \frac{{\sin A\left( {1 + \sin A} \right)}}{{\left( {1 + \sin A} \right)}} \cr & = 2 \cr} $$
86.
If 7sin2 ÃÂÃÂÃÂø + 3cos2 ÃÂÃÂÃÂø = 4, then the value of secÃÂÃÂÃÂø + cosecÃÂÃÂÃÂø is?
(A)
(B)
(C)
(D) None of these
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Solution:
$$\eqalign{ & {\text{7}}{\sin ^2}\theta + 3{\text{co}}{{\text{s}}^2}\theta = 4 \cr & {\text{7}}{\sin ^2}\theta + 3\left( {{\text{1}} - {\text{si}}{{\text{n}}^2}\theta } \right) = 4 \cr & {\text{7}}{\sin ^2}\theta + 3 - 3{\sin ^2}\theta = 4 \cr & 4{\sin ^2}\theta = 1 \cr & {\sin ^2}\theta = \frac{1}{4} \cr & {\text{sin }}\theta = \frac{1}{2} \cr & \sin \theta = \sin {30^ \circ } \cr & \theta = {30^ \circ } \cr & \sec \theta + \operatorname{cosec} \theta \cr & = \sec {30^ \circ } + \operatorname{cosec} {30^ \circ } \cr & = \frac{2}{{\sqrt 3 }} + 2 \cr} $$
87.
Let ÃÂÃÂ ÃÂÃÂ and ÃÂÃÂ Then a = b, if c = ?
(A) 1 - sinxcosx
(B) 1 + sinx - cosx
(C) 1 + sinxcosx
(D) 1 + cosx - sinx
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Solution:
$$\eqalign{ & b = \frac{c}{{1 + \sin x}} \cr & {\text{Go through option from option B}} \cr & b = \frac{{1 + \sin x - \cos x}}{{1 + \sin x}} \cr & = \frac{{{{\left( {1 + \sin x} \right)}^2} - {{\cos }^2}x}}{{1 + \sin x\left( {1 + \sin x + \cos x} \right)}} \cr & = \frac{{1 + {{\sin }^2}x + 2{{\sin }^2}x - 1 + {{\sin }^2}x}}{{\left( {1 + \sin x} \right)\left( {1 + \sin x + \cos x} \right)}} \cr & = \frac{{2{{\sin }^2}x + 2\sin x}}{{\left( {1 + \sin x} \right)\left( {1 + \sin x + \cos x} \right)}} \cr & = \frac{{2\sin x\left( {\sin x + 1} \right)}}{{1 + \sin x + \cos x}} \cr & = \frac{{2\sin x}}{{1 + \sin x + \cos x}} \cr & = a \cr} $$
88.
If ÃÂàÃÂàÃÂà0ÃÂÃÂÃÂð ÃÂÃÂÃÂø 90ÃÂÃÂÃÂð then the value of (tanÃÂÃÂÃÂø + secÃÂÃÂÃÂø)-1 is?
(A) 3 - √2
(B) 2 + √3
(C) 2 - √3
(D) 3 + √2
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Solution:
$$\eqalign{ & \frac{{\sin \theta }}{{1 + \cos \theta }} + \frac{{1 + \cos \theta }}{{\sin \theta }} = \frac{1}{{\sqrt 3 }} \cr & \frac{{{{\sin }^2}\theta + {{\left( {1 + \cos \theta } \right)}^2}}}{{\sin \theta \left( {1 + \cos \theta } \right)}} = \frac{1}{{\sqrt 3 }} \cr & \frac{{{{\sin }^2}\theta + 1 + {{\cos }^2}\theta + 2\cos \theta }}{{\sin \theta \left( {1 + \cos \theta } \right)}} = \frac{1}{{\sqrt 3 }} \cr & \frac{{{{\sin }^2}\theta + {{\cos }^2}\theta + 1 + 2\cos \theta }}{{\sin \theta \left( {1 + \cos \theta } \right)}} = \frac{1}{{\sqrt 3 }} \cr & \frac{{2\left( {1 + \cos \theta } \right)}}{{\sin \theta \left( {1 + \cos \theta } \right)}} = \frac{2}{{\sqrt 3 }} \cr & \sin \theta = \frac{{\sqrt 3 }}{2} \cr & \theta = {60^ \circ } \cr & \sec \theta - \tan \theta \cr & = \sec {60^ \circ } - \tan {60^ \circ } \cr & = 2 - \sqrt 3 \cr} $$
89.
ÃÂÃÂ ÃÂÃÂ is equal to:
(A) cotθ
(B) cosθsinθ
(C) secθcosecθ
(D) tanθ
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Solution:
$$\eqalign{ & \frac{{2 + {{\tan }^2}\theta + {{\cot }^2}\theta }}{{\sec \theta .{\text{cosec}}\,\theta }} \cr & = \frac{{{{\left( {\tan \theta + \cot \theta } \right)}^2}}}{{\sec \theta .{\text{cosec}}\,\theta }} \cr & = \frac{{{{\sin }^2}\theta + {{\cos }^2}\theta }}{{\left( {\sec \theta .{\text{cosec}}\,\theta } \right)\left( {{{\sin }^2}\theta .{{\cos }^2}\theta } \right)}} \cr & = \sec \theta .{\text{cosec}}\,\theta \cr} $$
90.
If ÃÂÃÂ ÃÂÃÂ the value of . ÃÂÃÂ = ?
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Solution:
$$\eqalign{ & {\bf{Shortcut\,\, method:}} \cr & {\text{Put }}\theta = {30^ \circ } \cr & \Rightarrow \sec \theta - \tan \theta = \frac{1}{{\sqrt 3 }} \cr & \Rightarrow \sec {30^ \circ } - \tan {30^ \circ } = \frac{1}{{\sqrt 3 }} \cr & \Rightarrow \frac{2}{{\sqrt 3 }} - \frac{1}{{\sqrt 3 }} = \frac{1}{{\sqrt 3 }} \cr & \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{1}{{\sqrt 3 }}{\text{ }}\left( {{\text{Satisfied}}} \right) \cr & \sec \theta = {30^ \circ } \cr & \Rightarrow \sec \theta .\tan \theta \cr & \Rightarrow \sec {30^ \circ }.\tan {30^ \circ } \cr & \Rightarrow \frac{2}{{\sqrt 3 }} \times \frac{1}{{\sqrt 3 }} \cr & \Rightarrow \frac{2}{3} \cr} $$