221.
(A) secθsinθ
(B) secθtanθ
(C) secθ + tanθ
(D) secθ - tanθ
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Solution:
Answer & Solution Answer: Option C No explanation is given for this question Let's Discuss on Board
222.
The value of the expression:
sin2 1ÃÂÃÂÃÂð + sin2 11ÃÂÃÂÃÂð + sin2 21ÃÂÃÂÃÂð + sin2 31ÃÂÃÂÃÂð + sin2 41ÃÂÃÂÃÂð + sin2 45ÃÂÃÂÃÂð + sin2 49ÃÂÃÂÃÂð + sin2 59ÃÂÃÂÃÂð + sin2 69ÃÂÃÂÃÂð + sin2 79ÃÂÃÂÃÂð + sin2 89ÃÂÃÂÃÂð is?
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Solution:
sin21° + sin211° + sin221° + sin231° + sin241° + sin245° + sin249° + sin259° + sin269° + sin279° + sin289° = (sin21° + sin289°) + (sin211° + sin279°) + (sin221° + sin269°) + (sin231° + sin259°) + (sin241° + sin249°) + sin245° = 1 + 1 + 1 + 1 + 1 + $$\frac{1}{2}$$ [sin2A + sin2B = 1. If, A + B = 90°] = $$5\frac{1}{2}$$
223.
If tan2 A + 2tanA - 63 = 0, given that 0 A what is the value of (2sinA + 5cosA)?
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Solution:
$$\eqalign{ & {\text{Given,}} \cr & {\tan ^2}A + 2\tan A - 63 = 0 \cr & {\bf{Formula \,used:}} \cr & {\text{Pythagoras Theorem,}} \cr & h = \sqrt {{p^2} + {b^2}} \cr & {\text{Where, h = Hypotenuse, p = Perpendicular and b = base}} \cr & {\bf{Calculation:}} \cr & {\tan ^2}A + 2\tan A - 63 = 0 \cr & {\text{Let, }}\tan A = x \cr & \Rightarrow {x^2} + 2x - 63 = 0 \cr & \Rightarrow {x^2} + 9x - 7x - 63 = 0 \cr & \Rightarrow x\left( {x + 9} \right) - 7\left( {x + 9} \right) = 0 \cr & \Rightarrow \left( {x + 9} \right)\left( {x - 7} \right) \cr & \Rightarrow x + 9 = 0 \Rightarrow x = - 9\left[ {'' - ''{\text{ will be neglected because }}0 A \frac{\pi }{2}} \right] \cr & \Rightarrow x - 7 = 0 \Rightarrow x = 7 \cr & {\text{So,}}\tan A = \frac{7}{1} = \frac{p}{b} \cr & {\text{Using Pythagoras Theorem,}} \cr & h = \sqrt {{p^2} + {b^2}} \cr & h = \sqrt {{7^2} + {1^2}} \cr & h = \sqrt {50} \cr & {\text{So}},\,2\sin A + 5\cos A \cr & = \left[ {2 \times \frac{7}{{\sqrt {50} }}} \right] + \left[ {5 \times \frac{1}{{\sqrt {50} }}} \right] \cr & = \frac{{14}}{{\sqrt {50} }} + \frac{5}{{\sqrt {50} }} \cr & = \frac{{14 + 5}}{{\sqrt {50} }} \cr & = \frac{{19}}{{\sqrt {50} }} \cr & \therefore {\text{The value of}}\left( {2\sin A + 5\cos A} \right){\text{is }}\frac{{19}}{{\sqrt {50} }}. \cr} $$
224.
If tanÃÂÃÂÃÂø + cotÃÂÃÂÃÂø = 2, then the value if tan100 ÃÂÃÂÃÂø + cot100 ÃÂÃÂÃÂø is?
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Solution:
$$\eqalign{ & {\text{tan}}\theta + {\text{cot}}\theta = 2 \cr & {\text{Put }}\theta = {45^ \circ } \cr & 1 + 1 = 2\left( {{\text{matched}}} \right) \cr & {\text{So, }}\theta = {45^ \circ } \cr & \Rightarrow {\text{ta}}{{\text{n}}^{100}}{45^ \circ } + {\text{co}}{{\text{t}}^{100}}{45^ \circ } \cr & \Rightarrow {1^{100}} + {1^{100}} \cr & \Rightarrow 2 \cr} $$
225.
The value of
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Solution:
$$\eqalign{ & \frac{{3\left( {{{\cot }^2}{{47}^ \circ } - {{\sec }^2}{{43}^ \circ }} \right) - 2\left( {{{\tan }^2}{{23}^ \circ } - {\text{cose}}{{\text{c}}^2}{{67}^ \circ }} \right)}}{{{\text{cose}}{{\text{c}}^2}\left( {{{68}^ \circ } + \theta } \right) - \tan \left( {\theta + {{61}^ \circ }} \right) - {{\tan }^2}\left( {{{22}^ \circ } - \theta } \right) + \cot \left( {{{29}^ \circ } - \theta } \right)}} \cr & = \frac{{3\left( {{{\tan }^2}{{43}^ \circ } - {{\sec }^2}{{43}^ \circ }} \right) - 2\left( {{{\tan }^2}{{23}^ \circ } - {{\sec }^2}{{23}^ \circ }} \right)}}{{{\text{cose}}{{\text{c}}^2}\left( {{{68}^ \circ } + \theta } \right) - \tan \left( {\theta + {{61}^ \circ }} \right) - {{\cot }^2}\left( {{{68}^ \circ } - \theta } \right) + \tan \left( {{{61}^ \circ } - \theta } \right)}} \cr & = \frac{{3 \times \left( { - 1} \right) - 2 \times \left( { - 1} \right)}}{1} \cr & = \frac{{ - 3 + 2}}{1} \cr & = - 1 \cr} $$
226.
The greatest value of sin4 ÃÂÃÂÃÂø + cos4 ÃÂÃÂÃÂø is?
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Solution:
$$\eqalign{ & {\sin ^2}\theta + {\cos ^2}\theta = 1 \cr & {\text{Squaring both sides}} \cr & {\sin ^4}\theta + {\cos ^4}\theta \cr & = 1 - 2{\sin ^2}\theta . {\cos ^2}\theta \cr & {\text{Put}}\,\theta = {90^ \circ } \cr & = 1 - 2{\sin ^2}{90^ \circ } \times {\cos ^2}{90^ \circ } \cr & = 1 - 0 \cr & = 1 \cr} $$
227.
The value of, ÃÂÃÂ ÃÂÃÂ ÃÂÃÂ - ÃÂÃÂ is?
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Solution:
$$\eqalign{ & {\text{sec}}\theta \left( {\frac{{1 + \sin \theta }}{{{\text{cos}}\theta }} + \frac{{{\text{cos}}\theta }}{{1 + \sin \theta }}} \right) - 2{\text{ta}}{{\text{n}}^2}\theta \cr & {\bf{Shortcut method:}} \cr & {\text{Take, }}\theta = {0^ \circ } \cr & \Rightarrow {\text{sec }}{0^ \circ }\left( {\frac{{1 + \sin {0^ \circ }}}{{{\text{cos }}{0^ \circ }}} + \frac{{{\text{cos }}{0^ \circ }}}{{1 + \sin {0^ \circ }}}} \right) - 2{\text{ta}}{{\text{n}}^2}{0^ \circ } \cr & \Rightarrow 1\left( {\frac{{1 + 0}}{1} + \frac{1}{{1 + 0}}} \right) - 0 \cr & \Rightarrow 2 \cr} $$
228.
If sec15ÃÂÃÂÃÂø = cosec15ÃÂÃÂÃÂø (0ÃÂÃÂÃÂð ÃÂÃÂÃÂø 10ÃÂÃÂÃÂð) then the value of ÃÂÃÂÃÂø is?
(A) 9°
(B) 5°
(C) 8°
(D) 3°
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Solution:
$$\eqalign{ & \sec 15\theta = \operatorname{cosec} 15\theta \cr & \Rightarrow \frac{1}{{\cos {{15} }\theta }} = \frac{1}{{\sin 15\theta }} \cr & \Rightarrow \frac{{\sin 15\theta }}{{{\text{cos15}}\theta }} = 1 \cr & \Rightarrow {\text{tan15}}\theta = 1 \cr & \Rightarrow {\text{tan15}}\theta = {\text{tan}}{45^ \circ } \cr & \Rightarrow 15\theta = {45^ \circ } \cr & \Rightarrow \theta = \frac{{45}}{{15}} \cr & \Rightarrow \theta = {3^ \circ } \cr} $$
229.
If sin(A + B) = cos(A + B), what is the value of tanA?
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Solution:
$$\eqalign{ & \sin \left( {A + B} \right) = \cos \left( {A + B} \right) \cr & \frac{{\sin \left( {A + B} \right)}}{{\cos \left( {A + B} \right)}} = 1 \cr & \tan \left( {A + B} \right) = 1 \cr & \tan \left( {A + B} \right) = \tan {45^ \circ } \cr & A + B = {45^ \circ } \cr & A = {45^ \circ } - B \cr & \tan A = \tan {45^ \circ } - \tan B \cr & \tan A = \frac{{\tan {{45}^ \circ } - \tan B}}{{1 + \tan {{45}^ \circ }\tan B}} \cr & \tan A = \frac{{1 - \tan B}}{{1 + \tan B}} \cr} $$
230.
If 3sinÃÂÃÂÃÂø = 2cos2 ÃÂÃÂÃÂø, 0ÃÂÃÂÃÂð ÃÂÃÂÃÂø 90ÃÂÃÂÃÂð, then the value of (tan2 ÃÂÃÂÃÂø + sec2 ÃÂÃÂÃÂø - cosec2 ÃÂÃÂÃÂø) is:
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Solution:
$$\eqalign{ & 3\sin \theta = 2{\cos ^2}\theta \cr & {\text{Let }}\theta = {30^ \circ } \cr & 3 \times \frac{1}{2} = 2 \times \frac{{{{\left( {\sqrt 3 } \right)}^2}}}{4} \cr & \frac{3}{2} = \frac{3}{2} \cr & {\tan ^2}\theta + {\sec ^2}\theta - {\text{cose}}{{\text{c}}^2}\theta \cr & = {\tan ^2}{30^ \circ } + {\sec ^2}{30^ \circ } - {\text{cose}}{{\text{c}}^2}{30^ \circ } \cr & = \frac{1}{3} + \frac{4}{3} - 4 \cr & = \frac{5}{3} - 4 \cr & = - \frac{7}{3} \cr} $$