141.
(sinÃÂÃÂÃÂø + cosecÃÂÃÂÃÂø)2 + (cosÃÂÃÂÃÂø + secÃÂÃÂÃÂø)2 - 1 = . . . . . . . .
(A) 6 + tan2θ + cot2θ
(B) tan2θ - cot2θ
(C) sin2θ - cot2θ
(D) 6 - sin2θ + cos2θ
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Solution:
(sinθ + cosecθ)2 + (cosθ + secθ)2 - 1 = sin2θ + cosec2θ + 2 + cos2θ + sec2θ + 2 - 1 = (sin2θ + cos2θ) + 1 + cot2θ + 2 + 1 + tan2θ + 2 - 1 = 6 + tan2θ + cot2θ
142.
The value of expression cos2 45ÃÂÃÂÃÂð + cos2 135ÃÂÃÂÃÂð + cos2 225ÃÂÃÂÃÂð + cos2 315ÃÂÃÂÃÂð is:
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Solution:
cos245° + cos2135° + cos2225° + cos2315° = (cos245° + cos2315°) + (cos2135° + cos2225°) [Here cos2α + cos2β = 1, where α + β = 360°] = 1 + 1 = 2
143.
The value of 4sin2 30ÃÂÃÂÃÂð + 3cot2 60ÃÂÃÂÃÂð is:
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Solution:
$$\eqalign{ & 4{\sin ^2}{30^ \circ } + 3{\cot ^2}{60^ \circ } \cr & = 4 \times {\left( {\frac{1}{2}} \right)^2} + 3{\left( {\frac{1}{{\sqrt 3 }}} \right)^2} \cr & = 4 \times \frac{1}{4} + 3 \times \frac{1}{3} \cr & = 1 + 1 \cr & = 2 \cr} $$
144.
The greatest value of sin4 ÃÂÃÂÃÂø + cos4 ÃÂÃÂÃÂø is?
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Solution:
$$\eqalign{ & {\sin ^2}\theta + {\cos ^2}\theta = 1 \cr & {\text{Squaring both sides}} \cr & {\sin ^4}\theta + {\cos ^4}\theta \cr & = 1 - 2{\sin ^2}\theta . {\cos ^2}\theta \cr & {\text{Put}}\,\theta = {90^ \circ } \cr & = 1 - 2{\sin ^2}{90^ \circ } \times {\cos ^2}{90^ \circ } \cr & = 1 - 0 \cr & = 1 \cr} $$
145.
Which of the following is equal to
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Solution:
$$\eqalign{ & \left[ {\frac{{\tan \theta + \sec \theta - 1}}{{\tan \theta - \sec \theta + 1}}} \right] \cr & = \left[ {\frac{{\tan \theta + \sec \theta - \left( {{{\sec }^2}\theta - {{\tan }^2}\theta } \right)}}{{\left( {\tan \theta - \sec \theta + 1} \right)}}} \right] \cr & = \frac{{\left( {\sec \theta + \tan \theta } \right)\left[ {1 - \sec \theta + \tan \theta } \right]}}{{\left( {\tan \theta - \sec \theta + 1} \right)}} \cr & = \sec \theta + \tan \theta \cr & = \frac{1}{{\cos \theta }} + \frac{{\sin \theta }}{{\cos \theta }} \cr & = \frac{{1 + \sin \theta }}{{\cos \theta }} \cr} $$
146.
The expression ÃÂÃÂ ÃÂÃÂ ÃÂÃÂ is equal to:
(A) -cos2θsin2θ
(B) sec2θcosec2θ
(C) -sec2θcosec2θ
(D) cos2θsin2θ
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Solution:
$$\eqalign{ & \frac{{{{\tan }^6}\theta - {{\sec }^6}\theta + 3{{\sec }^2}\theta \,{{\tan }^2}\theta }}{{{{\tan }^2}\theta + {{\cot }^2}\theta + 2}} \cr & = \frac{{{{\left( {{{\tan }^2}\theta } \right)}^3} - {{\left( {{{\sec }^2}\theta } \right)}^3} + 3{{\sec }^2}\theta \,{{\tan }^2}\theta }}{{{{\tan }^2}\theta + {{\cot }^2}\theta + 2}} \cr & \left[ {{a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)} \right] \cr & = \frac{{\left( {{{\tan }^2}\theta - {{\sec }^2}\theta } \right)\left( {{{\tan }^4}\theta + {{\sec }^4}\theta + {{\tan }^2}\theta {{\sec }^2}\theta + 3{{\sec }^2}\theta {{\tan }^2}\theta } \right)}}{{{{\tan }^2}\theta + {{\cot }^2}\theta + 2}} \cr & = \frac{{ - 1\left( {{{\tan }^4}\theta + {{\sec }^4}\theta + {{\tan }^2}\theta {{\sec }^2}\theta - 3{{\sec }^2}\theta {{\tan }^2}\theta } \right)}}{{{{\tan }^2}\theta + {{\cot }^2}\theta + 2}} \cr & = \frac{{ - 1\left( {{{\tan }^4}\theta + {{\sec }^4}\theta - 2{{\sec }^2}\theta {{\tan }^2}\theta } \right)}}{{{{\tan }^2}\theta + {{\cot }^2}\theta + 2}} \cr & = \frac{{ - 1\left( {{{\sec }^2}\theta - {{\tan }^2}\theta } \right)}}{{{{\tan }^2}\theta + {{\cot }^2}\theta + 2}} \cr & = \frac{{ - 1}}{{{{\left( {\tan \theta + \cot \theta } \right)}^2}}} \cr & = \frac{{ - 1}}{{{{\left( {\frac{{\sin \theta }}{{\cos \theta }} + \frac{{\cos \theta }}{{\sin \theta }}} \right)}^2}}} \cr & = \frac{{ - 1}}{{{{\left( {\frac{{{{\sin }^2}\theta + {{\cos }^2}\theta }}{{\cos \theta \sin \theta }}} \right)}^2}}} \cr & = - {\left( {\cos \theta \sin \theta } \right)^2} \cr & = - {\cos ^2}\theta {\sin ^2}\theta \cr} $$
147.
If ÃÂàÃÂà0ÃÂÃÂÃÂð ÃÂÃÂÃÂø 90ÃÂÃÂÃÂð, then the value of secÃÂÃÂÃÂø + tanÃÂÃÂÃÂø is:
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Solution:
$$\sin \theta = \frac{a}{{\sqrt {{a^2} + {b^2}} }}$$ $$\eqalign{ & \sec \theta + \tan \theta \cr & = \frac{{\sqrt {{a^2} + {b^2}} }}{b} + \frac{a}{b} \cr & = \frac{{\sqrt {{a^2} + {b^2}} + a}}{b} \cr} $$
148.
The value of sin2 65ÃÂÃÂÃÂð + sin2 25ÃÂÃÂÃÂð + cos2 35ÃÂÃÂÃÂð + cos2 55ÃÂÃÂÃÂð is?
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Solution:
$$\eqalign{ & \Rightarrow {\sin ^2}{65^ \circ } + {\sin ^2}{25^ \circ } + {\cos ^2}{35^ \circ } + {\cos ^2}{55^ \circ } \cr & \Rightarrow {\sin ^2}{65^ \circ } + {\sin ^2}\left( {{{90}^ \circ } - {{65}^ \circ }} \right) + \left[ {{{\cos }^2}{{35}^ \circ } + {{\cos }^2}\left( {{{90}^ \circ } - {{35}^ \circ }} \right)} \right] \cr & \Rightarrow \left( {{{\sin }^2}{{65}^ \circ } + {{\cos }^2}{{65}^ \circ }} \right) + \left( {{{\cos }^2}{{35}^ \circ } + si{n^2}{{35}^ \circ }} \right) \cr & \Rightarrow 1 + 1 \cr & \Rightarrow 2 \cr} $$
149.
A coconut tree swings with the wind in such a manner that the angle covered by its trunk is 18 degrees. If the topmost portion of the tree covers a distance of 44 metres, find the length of the tree.
(A) 120 metres
(B) 210 metres
(C) 140 metres
(D) 70 metres
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Solution:
$$\eqalign{ & {\text{Arc length}} = \frac{\theta }{{180}} \times \pi r \cr & \Rightarrow 44 = \frac{{18}}{{180}} \times \frac{{22}}{7} \times r \cr & \Rightarrow r = 140\,{\text{m}} \cr} $$
150.
If ÃÂÃÂ ÃÂÃÂ then ÃÂÃÂ + ÃÂÃÂ is equal to?
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Solution:
$$\eqalign{ & sec\theta = x + \frac{1}{{4x}} \cr & {\text{tan}}\theta = \sqrt {{\text{sec}}\theta - 1} \cr & {\text{tan}}\theta = \sqrt {{{\left[ {\frac{{4{x^2} + 1}}{{4x}}} \right]}^2} - 1} \cr & {\text{tan}}\theta = \sqrt {\frac{{{{\left( {4{x^2} + 1} \right)}^2} - {{\left( {4x} \right)}^2}}}{{{{\left( {4x} \right)}^2}}}} \cr & {\text{tan}}\theta = \sqrt {\frac{{16{x^4} + 1 + 8{x^2} - 16{x^2}}}{{{{\left( {4x} \right)}^2}}}} \cr & {\text{tan}}\theta = \sqrt {\frac{{16{x^4} + 1 - 8{x^2}}}{{{{\left( {4x} \right)}^2}}}} \cr & {\text{tan}}\theta = \sqrt {\frac{{{{\left( {4{x^2} - 1} \right)}^2}}}{{{{\left( {4x} \right)}^2}}}} \cr & {\text{tan}}\theta = \frac{{\left( {4{x^2} - 1} \right)}}{{4x}} \cr & \therefore sec\theta + {\text{tan}}\theta \cr & = \frac{{4{x^2} + 1}}{{4x}} + \frac{{4{x^2} - 1}}{{4x}} \cr & = \frac{{4{x^2} + 1 + 4{x^2} - 1}}{{4x}} \cr & = \frac{{8{x^2}}}{{4x}} \cr & = 2x \cr & \cr & {\bf{Alternate:}} \cr & sec\theta = x + \frac{1}{{4x}} \cr & {\text{Put }}x = 1 \cr & sec\theta = 1 + \frac{1}{4} = \frac{5}{4} = \frac{{\text{H}}}{{\text{B}}} \cr & \tan \theta = \frac{{\text{P}}}{{\text{B}}} = \frac{3}{4} \cr & {\text{Now, }} \cr & sec\theta + {\text{tan}}\theta \cr & = \frac{5}{4} + \frac{3}{4} \cr & = \frac{{5 + 3}}{4} \cr & = \frac{8}{4} \cr & = 2 \times 1 \cr & = 2x\left( {x = 1} \right) \cr} $$