21.
What is the value of cosec(65ÃÂÃÂÃÂð + ÃÂÃÂÃÂø) - sec(25ÃÂÃÂÃÂð - ÃÂÃÂÃÂø) + tan2 20ÃÂÃÂÃÂð - cosec2 70ÃÂÃÂÃÂð?
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Solution:
cosec(65° + θ) - sec(25° - θ) + tan220° - cosec270° = sec(25° - θ) - sec(25° - θ) + tan220° - sec220° = tan220° - sec220° = -1
22.
The value of
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Solution:
$$\eqalign{ & \frac{{\left( {\cos {9^ \circ } + \sin {{81}^ \circ }} \right)\left( {\sec {9^ \circ } + {\text{cosec}}\,{{81}^ \circ }} \right)}}{{\sin {{56}^ \circ }\sec {{34}^ \circ } + \cos {{25}^ \circ }{\text{cosec}}\,{{65}^ \circ }}} \cr & = \frac{{\left( {\cos {9^ \circ } + \cos {9^ \circ }} \right)\left( {\sec {9^ \circ } + \sec {9^ \circ }} \right)}}{{\cos {{34}^ \circ }\sec {{34}^ \circ } + \cos {{25}^ \circ }\sec {{25}^ \circ }}} \cr & = \frac{{2\cos {9^ \circ } \times 2\sec {9^ \circ }}}{{1 + 1}} \cr & = \frac{{4 \times \cos {9^ \circ }\sec {9^ \circ }}}{2} \cr & = 2 \cr} $$
23.
If 0ÃÂÃÂÃÂð ÃÂÃÂÃÂø 90ÃÂÃÂÃÂð and cos2 ÃÂÃÂÃÂø = 3(cot2 ÃÂÃÂÃÂø - cos2 ÃÂÃÂÃÂø) then the value of ÃÂàis:
(A) √3 + 2
(B) 2(2 - √3)
(C) 2(√3 - 1)
(D) √3 + 1
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Solution:
$$\eqalign{ & {\cos ^2}\theta = 3\left( {{{\cot }^2}\theta - {{\cos }^2}\theta } \right) \cr & {\cos ^2}\theta = 3{\cos ^2}\theta \left( {\frac{1}{{{{\sin }^2}\theta }} - 1} \right) \cr & 1 = 3\left( {{\text{cose}}{{\text{c}}^2}\theta - 1} \right) \cr & \frac{1}{3} + 1 = {\text{cose}}{{\text{c}}^2}\theta \cr & \frac{2}{{\sqrt 3 }} = {\text{cosec}}\,\theta \cr & {\text{cosec }}{60^ \circ } = {\text{cosec}}\,\theta \cr & \theta = {60^ \circ } \cr & \Rightarrow {\left( {\frac{1}{2}\sec \theta + \sin \theta } \right)^{ - 1}} \cr & = {\left( {\frac{1}{2}\sec {{60}^ \circ } + \sin {{60}^ \circ }} \right)^{ - 1}} \cr & = {\left( {\frac{1}{2} \times 2 + \frac{{\sqrt 3 }}{2}} \right)^{ - 1}} \cr & = {\left( {\frac{{2 + \sqrt 3 }}{2}} \right)^{ - 1}} \cr & = \frac{2}{{2 + \sqrt 3 }} \cr & = 2\left( {2 - \sqrt 3 } \right) \cr} $$
24.
In ÃÂÃÂÃÂÃÂÃÂÃÂABC, right-angled at B, AB = 7 cm and AC - BC = 1 cm. Find the value of sinC.
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Solution:
As we know triplet 7, 24, 25 AC - BC = 25 - 24 = 1 sinC = $$\frac{7}{{25}}$$
25.
The value of ÃÂÃÂ - ÃÂÃÂ - ÃÂÃÂ is?
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Solution:
$$\eqalign{ & = {\sec ^2}{17^ \circ } - \frac{1}{{{{\tan }^2}{{73}^ \circ }}} - \sin {17^ \circ }\sec {73^ \circ } \cr & = {\sec ^2}{17^ \circ } - {\cot ^2}{73^ \circ } - \sin {17^ \circ }\sec \left( {{{90}^ \circ } - {{17}^ \circ }} \right) \cr & = {\sec ^2}{17^ \circ } - {\cot ^2}\left( {{{90}^ \circ } - {{17}^ \circ }} \right) - \sin {17^ \circ }\cos ec{17^ \circ } \cr & = {\sec ^2}{17^ \circ } - {\tan ^2}{17^ \circ } - 1 \cr & = 1 - 1\left[ {\because {{\sec }^2}\theta - {{\tan }^2}\theta = 1} \right] \cr & = 0 \cr} $$
26.
If A, B and C be the angles of a triangle, the incorrect relation is ?
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Solution:
$$\eqalign{ & {\text{A}} + {\text{B}} + {\text{C}} = \pi = {\text{18}}{0^ \circ }{\text{ }} \cr & \Rightarrow \frac{{{\text{A + B}}}}{2} = \frac{{{{180}^ \circ }}}{2} - \frac{{\text{C}}}{2} \cr & \Rightarrow {\text{sin }}\left( {\frac{{{\text{A + B}}}}{2}} \right) \cr & \Rightarrow {\text{sin }}\left( {\frac{\pi }{2} - \frac{{\text{C}}}{2}} \right) \cr & \Rightarrow {\text{cos}}\frac{{\text{C}}}{2} \cr & \cr & {\bf{Similarly:}} \cr & {\text{cos }}\left( {\frac{{{\text{A}} + {\text{B}}}}{2}} \right) = {\text{sin}}\frac{{\text{C}}}{2} \cr & {\text{cot }}\left( {\frac{{{\text{A}} + {\text{B}}}}{2}} \right) = \tan \frac{{\text{C}}}{2} \cr & {\text{tan }}\left( {\frac{{{\text{A}} + {\text{B}}}}{2}} \right) = \cot \frac{{\text{C}}}{2} \cr & {\text{So, option C is incorrect}}{\text{.}} \cr} $$
27.
If ÃÂÃÂ ÃÂÃÂ lies in first quadrant, then the value of ÃÂÃÂ is:
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Solution:
$$\eqalign{ & \frac{{\sec \theta - \tan \theta }}{{\sec \theta + \tan \theta }} = \frac{1}{7} \cr & 7\sec \theta - 7\tan \theta = \sec \theta + \tan \theta \cr & 6\sec \theta = 8\tan \theta \cr & \sin \theta = \frac{6}{8} = \frac{3}{4} = \frac{P}{H} \cr & B = \sqrt {{4^2} - {3^2}} = \sqrt 7 \cr & \Rightarrow \frac{{{\text{cosec}}\,\theta + {{\cot }^2}\theta }}{{{\text{cosec}}\,\theta - {{\cot }^2}\theta }} \cr & = \frac{{\frac{H}{P} + {{\left( {\frac{B}{P}} \right)}^2}}}{{\frac{H}{P} - {{\left( {\frac{B}{P}} \right)}^2}}} \cr & = \frac{{\frac{4}{3} + {{\left( {\frac{{\sqrt 7 }}{3}} \right)}^2}}}{{\frac{4}{3} - {{\left( {\frac{{\sqrt 7 }}{3}} \right)}^2}}} \cr & = \frac{{\frac{4}{3} + \frac{7}{9}}}{{\frac{4}{3} - \frac{7}{9}}} \cr & = \frac{{\frac{{12 + 7}}{9}}}{{\frac{{12 - 7}}{9}}} \cr & = \frac{{19}}{5} \cr} $$
28.
What will be the value of sin10ÃÂÃÂÃÂð - sin3 10ÃÂÃÂÃÂð?
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Solution:
$$\eqalign{ & \sin 10 - \frac{4}{3}{\sin ^3}10 \cr & = \frac{{3\sin 10 - 4{{\sin }^3}10}}{3} \cr & = \frac{1}{3}\sin 3 \times {10^ \circ }\,\,\,\,\,\left[ {\because \sin A = 3\sin A - 4{{\sin }^3}A} \right] \cr & = \frac{1}{3} \times \frac{1}{2} \cr & = \frac{1}{6} \cr} $$
29.
The value of following is, cos24ÃÂÃÂÃÂð + cos55ÃÂÃÂÃÂð + cos125ÃÂÃÂÃÂð + cos204ÃÂÃÂÃÂð + cos300ÃÂÃÂÃÂð ?
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Solution:
The value of, cos24° + cos55° + cos125° + cos204° + cos300° We know that, cos(180° $$ \pm $$ θ) = -cosθ ⇒ cos24° + cos55° + cos (180° - 55°) + cos (180° + 24°) + cos (360° - 60°) ⇒ cos24° + cos55° - cos55° - cos24° + cos60° ⇒ cos60° ⇒ $$\frac{1}{2}$$
30.
If A = 10ÃÂÃÂÃÂð, what is the value of:
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Solution:
$$\eqalign{ & A = {10^ \circ } \cr & \frac{{12\sin 3A + 5\cos \left( {5A - {5^ \circ }} \right)}}{{9\sin \frac{{9A}}{2} - 4\cos \left( {5A + {{10}^ \circ }} \right)}} \cr & = \frac{{12\sin 3 \times {{10}^ \circ } + 5\cos \left( {5 \times {{10}^ \circ } - {5^ \circ }} \right)}}{{9\sin \frac{{9 \times {{10}^ \circ }}}{2} - 4\cos \left( {5 \times {{10}^ \circ } + {{10}^ \circ }} \right)}} \cr & = \frac{{12\sin {{30}^ \circ } + 5\cos \left( {{{50}^ \circ } - {5^ \circ }} \right)}}{{9\sin \frac{{{{90}^ \circ }}}{2} - 4\cos \left( {{{50}^ \circ } + {{10}^ \circ }} \right)}} \cr & = \frac{{12\sin {{30}^ \circ } + 5\cos {{45}^ \circ }}}{{9\sin {{45}^ \circ } - 4\cos {{60}^ \circ }}} \cr & = \frac{{12 \times \frac{1}{2} + 5 \times \frac{1}{{\sqrt 2 }}}}{{9 \times \frac{1}{{\sqrt 2 }} - 4 \times \frac{1}{2}}} \cr & = \frac{{6 + \frac{5}{{\sqrt 2 }}}}{{\frac{9}{{\sqrt 2 }} - 2}} \cr & = \frac{{6\sqrt 2 + 5}}{{9 - 2\sqrt 2 }} \cr} $$