331.
If ÃÂÃÂ + ÃÂÃÂ = 4, then the value of ÃÂÃÂ is?
- (A) 60°
- (B) 45°
- (C) 30°
- (D) 35°
Solution:
$$\eqalign{ & \Rightarrow \frac{{\cos \theta }}{{1 - \sin \theta }} + \frac{{\cos \theta }}{{1 + \sin \theta }} = 4 \cr & \Rightarrow \cos \theta \left( {\frac{{1 + \sin \theta + 1 - \sin \theta }}{{1 - {{\sin }^2}\theta }}} \right) = 4 \cr & \Rightarrow \cos \theta \left( {\frac{2}{{{\text{co}}{{\text{s}}^2}\theta }}} \right) = 4 \cr & \Rightarrow \cos \theta = \frac{1}{2} \cr & \Rightarrow \theta = {60^ \circ } \cr} $$
332.
The value of sec228ÃÂÃÂÃÂð - cot262ÃÂÃÂÃÂð + sin260ÃÂÃÂÃÂð + cosec230ÃÂÃÂÃÂð is equal to:
Solution:
$$\eqalign{ & {\sec ^2}{28^ \circ } - {\cot ^2}{62^ \circ } + {\sin ^2}{60^ \circ } + {\text{cose}}{{\text{c}}^2}{30^ \circ } \cr & = {\sec ^2}{28^ \circ } - {\tan ^2}{28^ \circ } + {\sin ^2}{60^ \circ } + {\text{cose}}{{\text{c}}^2}{30^ \circ } \cr & = 1 + {\left( {\frac{{\sqrt 3 }}{2}} \right)^2} + {\left( 2 \right)^2} \cr & = 1 + \frac{3}{4} + 4 \cr & = \frac{{23}}{4} \cr} $$
333.
If ÃÂÃÂ ÃÂÃÂ then ÃÂÃÂ + ÃÂÃÂ is equal to?
Solution:
$$\eqalign{ & sec\theta = x + \frac{1}{{4x}} \cr & {\text{tan}}\theta = \sqrt {{\text{sec}}\theta - 1} \cr & {\text{tan}}\theta = \sqrt {{{\left[ {\frac{{4{x^2} + 1}}{{4x}}} \right]}^2} - 1} \cr & {\text{tan}}\theta = \sqrt {\frac{{{{\left( {4{x^2} + 1} \right)}^2} - {{\left( {4x} \right)}^2}}}{{{{\left( {4x} \right)}^2}}}} \cr & {\text{tan}}\theta = \sqrt {\frac{{16{x^4} + 1 + 8{x^2} - 16{x^2}}}{{{{\left( {4x} \right)}^2}}}} \cr & {\text{tan}}\theta = \sqrt {\frac{{16{x^4} + 1 - 8{x^2}}}{{{{\left( {4x} \right)}^2}}}} \cr & {\text{tan}}\theta = \sqrt {\frac{{{{\left( {4{x^2} - 1} \right)}^2}}}{{{{\left( {4x} \right)}^2}}}} \cr & {\text{tan}}\theta = \frac{{\left( {4{x^2} - 1} \right)}}{{4x}} \cr & \therefore sec\theta + {\text{tan}}\theta \cr & = \frac{{4{x^2} + 1}}{{4x}} + \frac{{4{x^2} - 1}}{{4x}} \cr & = \frac{{4{x^2} + 1 + 4{x^2} - 1}}{{4x}} \cr & = \frac{{8{x^2}}}{{4x}} \cr & = 2x \cr & \cr & {\bf{Alternate:}} \cr & sec\theta = x + \frac{1}{{4x}} \cr & {\text{Put }}x = 1 \cr & sec\theta = 1 + \frac{1}{4} = \frac{5}{4} = \frac{{\text{H}}}{{\text{B}}} \cr & \tan \theta = \frac{{\text{P}}}{{\text{B}}} = \frac{3}{4} \cr & {\text{Now, }} \cr & sec\theta + {\text{tan}}\theta \cr & = \frac{5}{4} + \frac{3}{4} \cr & = \frac{{5 + 3}}{4} \cr & = \frac{8}{4} \cr & = 2 \times 1 \cr & = 2x\left( {x = 1} \right) \cr} $$
334.
If ÃÂÃÂ then the value of ÃÂÃÂ is?
Solution:
$$\eqalign{ & \cot \theta = 4 \cr & \therefore \frac{{5\sin \theta + 3\cos \theta }}{{5\sin \theta - 3\cos \theta }} \cr & = \frac{{5 + 3\cot \theta }}{{5 - 3\cot \theta }} \cr & = \frac{{5 + 3 \times 4}}{{5 - 3 \times 4}} \cr & = - \frac{{17}}{7} \cr} $$
335.
The value of tan 4ÃÂÃÂÃÂð.tan 43ÃÂÃÂÃÂð.tan 47ÃÂÃÂÃÂð.tan 86ÃÂÃÂÃÂð is?
Solution:
$$\eqalign{ & {\text{tan }}{{\text{4}}^ \circ }{\text{.tan 4}}{{\text{3}}^ \circ }{\text{.tan 4}}{{\text{7}}^ \circ }{\text{.tan 8}}{{\text{6}}^ \circ } \cr & {\text{Here, }} \cr & {\text{tan 8}}{{\text{6}}^ \circ } = {\text{tan}}\left( {{\text{ 9}}{{\text{0}}^ \circ } - {4^ \circ }} \right) = {\text{cot }}{{\text{4}}^ \circ } \cr & {\text{tan 4}}{{\text{7}}^ \circ } = {\text{tan}}\left( {{\text{ 9}}{{\text{0}}^ \circ } - {{43}^ \circ }} \right) = {\text{cot 4}}{{\text{3}}^ \circ } \cr & {\text{tan }}{{\text{4}}^ \circ }{\text{.cot }}{{\text{4}}^ \circ }{\text{.tan 4}}{{\text{3}}^ \circ }{\text{.cot 4}}{{\text{3}}^ \circ } = 1 \cr} $$
336.
The value of sec228ÃÂÃÂÃÂð - cot262ÃÂÃÂÃÂð + sin260ÃÂÃÂÃÂð + cosec230ÃÂÃÂÃÂð is equal to:
Solution:
$$\eqalign{ & {\sec ^2}{28^ \circ } - {\cot ^2}{62^ \circ } + {\sin ^2}{60^ \circ } + {\text{cose}}{{\text{c}}^2}{30^ \circ } \cr & = {\sec ^2}{28^ \circ } - {\tan ^2}{28^ \circ } + {\sin ^2}{60^ \circ } + {\text{cose}}{{\text{c}}^2}{30^ \circ } \cr & = 1 + {\left( {\frac{{\sqrt 3 }}{2}} \right)^2} + {\left( 2 \right)^2} \cr & = 1 + \frac{3}{4} + 4 \cr & = \frac{{23}}{4} \cr} $$
337.
The value of
- (A) secθ
- (B) sinθ
- (C) cotθ
- (D) tanθ
Solution:
$$\eqalign{ & \frac{{2{{\cos }^3}\theta - \cos \theta }}{{\sin \theta - 2{{\sin }^3}\theta }} \cr & = \frac{{\cos \theta \left[ {2{{\cos }^2}\theta - 1} \right]}}{{\sin \theta \left[ {1 - 2{{\sin }^2}\theta } \right]}} \cr & = \frac{{\cos \theta \times \cos 2\theta }}{{\sin \theta \times \cos 2\theta }} \cr & = \cot \theta \cr} $$
338.
If tanÃÂÃÂÃÂø + cotÃÂÃÂÃÂø = -2, then the value of tan9ÃÂÃÂÃÂø + cot9ÃÂÃÂÃÂø is:
- (A) -2
- (B) 0
- (C) 2
- (D) -1
Solution:
$$\eqalign{ & \tan \theta + \cot \theta = - 2 \cr & \Rightarrow \tan \theta + \frac{1}{{\tan \theta }} = - 2 \cr & \Rightarrow \tan \theta = - 1 \cr & \Rightarrow {\tan ^9}\theta + {\cot ^9}\theta \cr & = {\tan ^9}\theta + \frac{1}{{{{\tan }^9}\theta }} \cr & = {\left( { - 1} \right)^9} + \frac{1}{{{{\left( { - 1} \right)}^9}}} \cr & = - 1 - 1 \cr & = - 2 \cr} $$
339.
If 1 + cot2ÃÂÃÂÃÂø = ÃÂàand ÃÂÃÂÃÂø is acute, then what is the value of
Solution:
$$\eqalign{ & 1 + {\cot ^2}\theta = \frac{{625}}{{49}} \cr & {\text{cose}}{{\text{c}}^2}\theta = \frac{{625}}{{49}} \cr & {\text{cosec}}\,\theta = \frac{{25}}{7} \cr & \sin \theta = \frac{7}{{25}} \cr & \cos \theta = \frac{{24}}{{25}} \cr & {\left( {\sin \theta + \cos \theta } \right)^{\frac{1}{2}}} \cr & = {\left( {\frac{7}{{25}} + \frac{{24}}{{25}}} \right)^{\frac{1}{2}}} \cr & = {\left( {\frac{{31}}{{25}}} \right)^{\frac{1}{2}}} \cr & = \frac{{\sqrt {31} }}{5} \cr} $$
340.
What is
Solution:
$$\sin \alpha - \sin \beta = 2\cos \frac{{\alpha + \beta }}{2}.\sin \frac{{\alpha - \beta }}{2}$$