51.
Find the value of, 8cos10ÃÂÃÂÃÂð. cos20ÃÂÃÂÃÂð. cos40ÃÂÃÂÃÂð = ?
(A) 2cot20°
(B) 4tan10°
(C) 1
(D) cot10°
Show Answer
Solution:
Let x = 8cos10°. cos20°. cos40° Multiply on both side by sin10° and applying formula (2sinθ. cosθ = sin2θ) ⇒ x sin10° = 4 × 2sin10° cos10°. cos20°. cos40° ⇒ x sin10° = 2 × 2sin20°.cos20°. cos40° ⇒ x sin10° = 2 × sin40°. cos40° ⇒ x sin10° = sin80° ⇒ x sin10° = sin(90° - 10°) ⇒ x sin10° = cos10° then, x = $$\frac{{\cos {{10}^ \circ }}}{{\sin {{10}^ \circ }}}$$ x = cot10°
52.
(A) cosθ(1 + sinθ)
(B) secθ(1 + sinθ)
(C) 2cosθ(1 + secθ)
(D) 2secθ(1 + secθ)
Show Answer
Solution:
$$\eqalign{ & \frac{{{{\left( {1 + \cos \theta } \right)}^2} + {{\sin }^2}\theta }}{{\left( {{\text{cose}}{{\text{c}}^2}\theta - 1} \right){{\sin }^2}\theta }} \cr & = \frac{{1 + {{\cos }^2}\theta + 2\cos \theta + {{\sin }^2}\theta }}{{\left( {{\text{cose}}{{\text{c}}^2}\theta - 1} \right){{\sin }^2}\theta }} \cr & = \frac{{2\left( {1 + \cos \theta } \right)}}{{\frac{{\left( {1 - {{\sin }^2}\theta } \right)}}{{{{\sin }^2}\theta }}.{{\sin }^2}\theta }} \cr & = \frac{{2\left( {\cos \theta + 1} \right)}}{{{{\cos }^2}\theta }} \cr & = 2\sec \theta \left( {\frac{{\cos \theta }}{{\cos \theta }} + \frac{1}{{\cos \theta }}} \right) \cr & = 2\sec \theta \left( {1 + \sec \theta } \right) \cr} $$
53.
ABCD is a rectangle of which AC is a diagonal. The value of (tan ÃÂÃÂÃÂâÃÂÃÂÃÂÃÂÃÂàCAD + 1)sin ÃÂÃÂÃÂâÃÂÃÂÃÂÃÂÃÂàBAC is?22
Show Answer
Solution:
= (tan2α + 1 )sin2β = (tan245° + 1 )sin245° $$ = (1 + 1) {\left( {\frac{1}{{\sqrt 2 }}} \right)^2}$$ $$ = 2 \times \frac{1}{2} = 1$$
54.
The value of sin2 65ÃÂÃÂÃÂð + sin2 25ÃÂÃÂÃÂð + cos2 35ÃÂÃÂÃÂð + cos2 55ÃÂÃÂÃÂð is?
Show Answer
Solution:
$$\eqalign{ & \Rightarrow {\sin ^2}{65^ \circ } + {\sin ^2}{25^ \circ } + {\cos ^2}{35^ \circ } + {\cos ^2}{55^ \circ } \cr & \Rightarrow {\sin ^2}{65^ \circ } + {\sin ^2}\left( {{{90}^ \circ } - {{65}^ \circ }} \right) + \left[ {{{\cos }^2}{{35}^ \circ } + {{\cos }^2}\left( {{{90}^ \circ } - {{35}^ \circ }} \right)} \right] \cr & \Rightarrow \left( {{{\sin }^2}{{65}^ \circ } + {{\cos }^2}{{65}^ \circ }} \right) + \left( {{{\cos }^2}{{35}^ \circ } + si{n^2}{{35}^ \circ }} \right) \cr & \Rightarrow 1 + 1 \cr & \Rightarrow 2 \cr} $$
55.
If 3sinÃÂÃÂÃÂø = 2cos2 ÃÂÃÂÃÂø, 0ÃÂÃÂÃÂð ÃÂÃÂÃÂø 90ÃÂÃÂÃÂð, then the value of (tan2 ÃÂÃÂÃÂø + sec2 ÃÂÃÂÃÂø - cosec2 ÃÂÃÂÃÂø) is:
Show Answer
Solution:
$$\eqalign{ & 3\sin \theta = 2{\cos ^2}\theta \cr & {\text{Let }}\theta = {30^ \circ } \cr & 3 \times \frac{1}{2} = 2 \times \frac{{{{\left( {\sqrt 3 } \right)}^2}}}{4} \cr & \frac{3}{2} = \frac{3}{2} \cr & {\tan ^2}\theta + {\sec ^2}\theta - {\text{cose}}{{\text{c}}^2}\theta \cr & = {\tan ^2}{30^ \circ } + {\sec ^2}{30^ \circ } - {\text{cose}}{{\text{c}}^2}{30^ \circ } \cr & = \frac{1}{3} + \frac{4}{3} - 4 \cr & = \frac{5}{3} - 4 \cr & = - \frac{7}{3} \cr} $$
56.
If ÃÂÃÂ + ÃÂÃÂ = 4, then the value of ÃÂÃÂ is?
(A) 60°
(B) 45°
(C) 30°
(D) 35°
Show Answer
Solution:
$$\eqalign{ & \Rightarrow \frac{{\cos \theta }}{{1 - \sin \theta }} + \frac{{\cos \theta }}{{1 + \sin \theta }} = 4 \cr & \Rightarrow \cos \theta \left( {\frac{{1 + \sin \theta + 1 - \sin \theta }}{{1 - {{\sin }^2}\theta }}} \right) = 4 \cr & \Rightarrow \cos \theta \left( {\frac{2}{{{\text{co}}{{\text{s}}^2}\theta }}} \right) = 4 \cr & \Rightarrow \cos \theta = \frac{1}{2} \cr & \Rightarrow \theta = {60^ \circ } \cr} $$
57.
(secÃÂÃÂÃÂâÃÂÃÂÃÂÃÂ
- tanÃÂÃÂÃÂâÃÂÃÂÃÂÃÂ
)(1 + sinÃÂÃÂÃÂâÃÂÃÂÃÂÃÂ
) ÃÂÃÂÃÂÃÂÃÂÃÂÃÂ÷ cosÃÂÃÂÃÂâÃÂÃÂÃÂÃÂ
= ?222
(A) cos2∅
(B) 1
(C) cot2∅
(D) -1
Show Answer
Solution:
$$\eqalign{ & {\left( {\sec \phi - \tan \phi } \right)^2}{\left( {1 + \sin \phi } \right)^2} \div {\cos ^2}\phi \cr & = {\left( {\frac{1}{{\cos \phi }} - \frac{{\sin \phi }}{{\cos \phi }}} \right)^2}{\left( {1 + \sin \phi } \right)^2} \div {\cos ^2}\phi \cr & = {\left( {\frac{{1 - \sin \phi }}{{\cos \phi }}} \right)^2}{\left( {1 + \sin \phi } \right)^2} \div {\cos ^2}\phi \cr & = \frac{{{{\left( {1 - \sin \phi } \right)}^2}{{\left( {1 + \sin \phi } \right)}^2}}}{{{{\cos }^2}\phi }} \times \frac{1}{{{{\cos }^2}\phi }} \cr & = \frac{{{{\left( {1 - {{\sin }^2}\phi } \right)}^2}}}{{{{\cos }^4}\phi }} \cr & = \frac{{{{\cos }^4}\phi }}{{{{\cos }^4}\phi }} \cr & = 1 \cr} $$
58.
If sinÃÂÃÂÃÂø = 4cosÃÂÃÂÃÂø, then what is the value of sinÃÂÃÂÃÂøcosÃÂÃÂÃÂø?
Show Answer
Solution:
$$\eqalign{ & \sin \theta = 4\cos \theta \cr & \tan \theta = 4 \cr & \sin \theta .\cos \theta = \frac{4}{{\sqrt {17} }} \times \frac{1}{{\sqrt {17} }} = \frac{4}{{17}} \cr} $$
59.
If ÃÂÃÂÃÂÃÂÃÂÃÂÃÂñ + ÃÂÃÂÃÂÃÂÃÂÃÂÃÂò = 90ÃÂÃÂÃÂÃÂÃÂÃÂÃÂð, then the value of (1 - sinÃÂÃÂÃÂÃÂÃÂÃÂÃÂñ)(1 - cosÃÂÃÂÃÂÃÂÃÂÃÂÃÂñ) ÃÂÃÂÃÂÃÂÃÂà(1 + cotÃÂÃÂÃÂÃÂÃÂÃÂÃÂò)(1 + tanÃÂÃÂÃÂÃÂÃÂÃÂÃÂò) is?2222
Show Answer
Solution:
Shortcut method : $$\left( {1 - {{\sin }^2}\alpha } \right)\left( {1 - {{\cos }^2}\alpha } \right) \times $$ $$\left( {1 + {{\cot }^2}\beta } \right)$$ $$\left( {1 + {{\tan }^2}\beta } \right)$$ $$\eqalign{ & \Rightarrow \left( {{{\cos }^2}\alpha } \right)\left( {si{n^2}\alpha } \right) \left( {{{\operatorname{cosec} }^2}\beta } \right)\left( {{{\sec }^2}\beta } \right) \cr & {\text{Put }} \alpha = \beta = {45^ \circ } \cr & \Rightarrow {\text{co}}{{\text{s}}^2}{45^ \circ }.{\sin ^2}{45^ \circ }.{\text{cose}}{{\text{c}}^2}{45^ \circ }.{\operatorname{sce} ^2}{45^ \circ } \cr & \Rightarrow \frac{1}{2}.\frac{1}{2}.2.2 \cr & \Rightarrow 1 \cr} $$ Alternate : $$\left( {1 - {{\sin }^2}\alpha } \right)\left( {1 - {{\cos }^2}\alpha } \right) \times $$ $$\left( {1 + {{\cot }^2}\beta } \right)$$ $$\left( {1 + {{\tan }^2}\beta } \right)$$ $$\eqalign{ & \Rightarrow \left( {{{\cos }^2}\alpha } \right)\left( {si{n^2}\alpha } \right) \times \left( {{{\operatorname{cosec} }^2}\beta } \right)\left( {{{\sec }^2}\beta } \right) \cr & \Rightarrow {\text{co}}{{\text{s}}^2}\left( {{{90}^ \circ } - \beta } \right).{\sin ^2}\alpha .{\text{cose}}{{\text{c}}^2}\beta .{\text{se}}{{\text{c}}^2}\left( {{{90}^ \circ } - \alpha } \right) \cr & \Rightarrow {\sin ^2}\beta .{\text{cose}}{{\text{c}}^2}\beta .{\sin ^2}\alpha . {\text{cose}}{{\text{c}} ^2}\alpha \cr & \Rightarrow 1 \cr} $$
60.
If cosÃÂÃÂÃÂø + sinÃÂÃÂÃÂø = cosÃÂÃÂÃÂø, then cosÃÂÃÂÃÂø - sinÃÂÃÂÃÂø is?
(A) tanθ
(B) - cosθ
(C) - sinθ
(D) sinθ
Show Answer
Solution:
$$\eqalign{ & {\text{cos}}\theta + \sin \theta = \sqrt 2 \cos \theta \cr & {\text{Squaring both sides}} \cr & {\text{co}}{{\text{s}}^2}\theta + {\sin ^2}\theta + 2\cos \theta .\sin \theta = 2{\text{co}}{{\text{s}}^2}\theta \cr & \Rightarrow 2{\text{co}}{{\text{s}}^2}\theta - {\text{co}}{{\text{s}}^2}\theta - {\sin ^2}\theta = 2{\text{cos}}\theta .{\text{sin}}\theta \cr & \Rightarrow {\text{co}}{{\text{s}}^2}\theta - {\sin ^2}\theta = 2\sin \theta .{\text{cos}}\theta \cr & \Rightarrow \left( {\cos \theta - \sin \theta } \right)\left( {\cos \theta + \sin \theta } \right) = 2\sin \theta .{\text{cos}}\theta \cr & \Rightarrow \left( {\cos \theta - \sin \theta } \right)\left( {\sqrt 2 \cos \theta } \right) = 2\sin \theta .{\text{cos}}\theta \cr & \Rightarrow \cos \theta - \sin \theta = \frac{{2\sin \theta .\cos \theta }}{{\sqrt 2 \cos \theta }} \cr & \Rightarrow \sqrt 2 \sin \theta \cr & \cr & {\bf{Alternate:}} \cr & {\text{Let }}\sqrt 2 \cos \theta = \alpha \cr & \therefore \cos \theta \pm \sin \theta = a \cr & \cos \theta \pm \sin \theta = \sqrt {2 - {a^2}} \cr & = \sqrt {2 - {a^2}} \cr & = \sqrt {2 - 2{\text{co}}{{\text{s}}^2}\theta } \cr & = \sqrt {2\left( {1 - {\text{co}}{{\text{s}}^2}\theta } \right)} \cr & = \sqrt {2{{\sin }^2}\theta } \cr & = \sqrt 2 \sin \theta \cr} $$