01. if\ x\ sin^{2} 60\degree -\frac{3}{2} sec60\degree \ tan^{2} 30\degree +\frac{4}{5} sin^{2} 45\degree tan^{2} 60\degree =0\\ \\ then\ tan\ x\ is\ ?\ \\\ \\
( a) -\frac{1}{15} \\\ \\
( b) -4 \\\ \\
( c) -\frac{4}{15} \\\ \\
( d) -2 \\ \\
Read Solution
\Longrightarrow x\ sin^{2} 60\degree -\frac{3}{2} sec60\degree \ tan^{2} 30\degree +\frac{4}{5} sin^{2} 45\degree tan^{2} 60\degree =0 \\\ \\ ( we\ know\ sin60\degree =\frac{\sqrt{3}}{2} ,\ sec60\degree =2,\\ \\ tan30\degree =\frac{1}{\sqrt{3}} ,\ sin45\degree =\frac{1}{\sqrt{2}} \\ \\ \&\ tan60\degree =\sqrt{3}) \\\ \\ \Longrightarrow x\left(\frac{\sqrt{3}}{2}\right)^{2} -\frac{3}{2} \times 2\times \left(\frac{1}{\sqrt{3}}\right)^{2} +\frac{4}{5}\left(\frac{1}{\sqrt{2}}\right)^{2} \times \left(\sqrt{3}\right)^{2} =0 \\\ \\ \Longrightarrow \frac{3x}{4} -\left(\frac{3}{2} \times 2\times \frac{1}{3}\right) +\left(\frac{4}{5} \times \frac{1}{2} \times 3\right) =0 \\\ \\ \Longrightarrow \frac{3x}{4} -1+\frac{6}{5} =0 \\\ \\ \Longrightarrow \frac{15x-20+24}{20} =0\\\ \\ \Longrightarrow 15x+4=0\\\ \\ \Longrightarrow 15x=-4\\\ \\ \Longrightarrow x=-\frac{4}{15}\\\ \\
Option (c) is the right answer
02. if\ tan( A+B) =\sqrt{3} \\ \\ and\ tan( A-B) =\frac{1}{\sqrt{3}}, \\ \\ ( \angle A+\angle B) < 90\degree ,A\geqslant B\\ \\ find\ tan\ \angle A\ \\\ \\
( a) 90\degree \\\ \\
( b) 30\degree \\\ \\
( c) 45\degree \\\ \\
( d) 60\degree \\ \\
Read Solution
given\ tan( A+B) =\sqrt{3} \ \&\ tan( A-B) =\frac{1}{\sqrt{3}}\\\ \\ \Longrightarrow tan( A+B) =\sqrt{3}\\\ \\ \left( we\ know\ tan60\degree =\sqrt{3}\right)\\\ \\ \Longrightarrow tan( A+B) =tan60\degree \\\ \\ \Longrightarrow A+B=60\degree .....equation\ no1\\\ \\ \\\ \\ \Longrightarrow \ tan( A-B) =\frac{1}{\sqrt{3}}\\\ \\ \left( we\ know\ tan30\degree =\frac{1}{\sqrt{3}}\right)\\\ \\ \Longrightarrow \ tan( A-B) =tan30\degree \\\ \\ \Longrightarrow A-B=30\degree \ \ ....equation\ no2\\\ \\ \\\ \\ now\ equ.1\ \&\ equ.2\ adding\\\ \\ \Longrightarrow ( A+B) +( A-B) =60\degree +30\degree \\\ \\ \Longrightarrow 2A=90\degree \\\ \\ \Longrightarrow A=45\degree \\\ \\ ( \ put\ A=45\degree \ in\ equ.\ no\ 1)\\\ \\ \Longrightarrow A+B=60\degree \\\ \\ \Longrightarrow 45\degree +B=60\degree \\\ \\ \Longrightarrow B=60\degree -45\degree \\\ \\ B=15\degree \\\ \\ by\ question\\\ \\ ( \angle A+\angle B) < 90\degree ,A\geqslant B\\\ \\ 60\degree +15\degree < 90\degree ,\ 60\degree \geqslant 15\degree \ proved\\\ \\ so\ \angle A=45\degree \\\ \\
Option (c) is the right answer
03. if\ tanA+cotA=2\\ \\ Find\ value\ of\ tan^{10} A+cot^{10} A\ \\\ \\
(a) 4 \\\ \\
(b) 2 \\\ \\
(c) 2^{10} \\\ \\
(d) 1 \\ \\
Read Solution
\Longrightarrow tanA+cotA=2 \\\ \\ \left( we\ know\ cot\theta =\frac{1}{tan\theta }\right)\\\ \\ \Longrightarrow tanA+\frac{1}{tanA} =2\\\ \\ \Longrightarrow tan^{2} A+1=2tanA \\\ \\ \Longrightarrow tan^{2} A+1-2tanA=0\ \\\ \\ \left[ use\ formula\ ( a-b)^{2} =a^{2} -2ab+b^{2}\right]\\\ \\ \Longrightarrow ( tanA-1)^{2} =0\\\ \\ \Longrightarrow tanA-1=0\\\ \\ \Longrightarrow tanA=1\\\ \\ \left( we\ know\ tan\theta =\frac{1}{cot\theta }\right) \\\ \\ \Longrightarrow \frac{1}{cotA} =1 \\\ \\ \Longrightarrow cotA=1 \\\ \\ ( \ put\ the\ value\ tanA=1\ \&\ cotA=1)\\\ \\ =tan^{10} A+cot^{10} A\\\ \\ =1^{10} +1^{10}\\\ \\ =1+1 \\\ \\ =2\\\ \\
option (b) is the right answer
04. if\ xcos\theta -sin\theta =1,\\ \\ than\ x^{2} -\left( 1+x^{2}\right) sin\theta \ is\ equal\\\ \\
(a) 1 \\\ \\
(b) -1 \\\ \\
(c) 0 \\\ \\
(d) 2 \\ \\
Read Solution
given\ xcos\theta -sin\theta =1 \\\ \\ \Longrightarrow xcos\theta -sin\theta =1 \\\ \\ \Longrightarrow xcos\theta =1-sin\theta \\\ \\ let\ \theta =0\degree \\\ \\ \Longrightarrow xcos0\degree =1-sin0\degree \\\ \\ \Longrightarrow x\times 1=1-0\\\ \\ \Longrightarrow x=1 \\\ \\ ( put\ the\ value\ x=1)\\\ \\ =x^{2} -\left( 1+x^{2}\right) sin\theta \\\ \\ =1^{2} -\left( 1+1^{2}\right) sin0\degree \\\ \\ =1-2\times 0\\\ \\ =1\\\ \\
Option (a) is the right answer
05. Find \ value\ of\\ \\ sec^{4} A\left( 1-sin^{4} A\right) -2tan^{2} A\ \\\ \\
(a)\frac{1}{2} \\\ \\
(b) 0 \\\ \\
(c) 2 \\\ \\
(d) 1 \\ \\
Read Solution
=sec^{4} A\left( 1-sin^{4} A\right) -2tan^{2} A \\\ \\ =sec^{4} A-sec^{4} A.sin^{4} A-2tan^{2} A\\\ \\ =sec^{4} A-\frac{1}{cos^{4} A} .sin^{4} A-2tan^{2} A\\\ \\ \left( we\ know\ sec\theta =\frac{1}{cos\theta } \ \&\ tan\theta =\frac{sin\theta }{cos\theta }\right)\\\ \\ =sec^{4} A-tan^{4} A-2tan^{2} A \\\ \\ =\left( sec^{2} A\right)^{2} -\left( tan^{2} A\right)^{2} -2tan^{2} A\\\ \\ \left[ use\ formula\ a^{2} -b^{2} =( a-b)( a+b)\right] \\\ \\ =\left( sec^{2} A-tan^{2} A\right)\left( sec^{2} A+tan^{2} A\right) -2tan^{2} A\\\ \\ \ \left( we\ know\ sec^{2} A-tan^{2} A=1\right)\\\ \\ =sec^{2} A+tan^{2} A-2tan^{2} A\\\ \\ =sec^{2} A+tan^{2} A\\\ \\ =1 \\\ \\
Option (d) is the right answer
(06) \ if\ sin2\theta =\frac{\sqrt{3}}{2} \\ \\ than\ the\ value\ of\ sin3\theta \ is\ equal\ to\ ?\\ \\ take\ ( 0\degree \leqslant \theta \leqslant 90\degree )\\\ \\
(a)\frac{1}{2} \\ \\
(b)\frac{\sqrt{3}}{2} \\ \\
(c) 0 \\ \\
(d) 1\\ \\
Read Solution
\Longrightarrow sin2\theta =\frac{\sqrt{3}}{2}\\\ \\ (we\ know\ sin60\degree =\frac{\sqrt{3}}{2} )\\\ \\ \Longrightarrow sin2\theta =sin60\degree \\\ \\ \Longrightarrow 2\theta =60\degree \\\ \\ \Longrightarrow \theta =30\degree \\\ \\ \\\ \\ now\\\ \\ =sin3\theta \\\ \\ ( put\ the\ value\ \theta =30\degree )\\\ \\ =sin( 3\times 30)\\\ \\ =sin90\degree \\\ \\ =1 \\\ \\
Option (d) is the right answer
(07) if\ \theta \ be\ an\ acute\ angle\ and\\ \\ tan\theta +cot\theta =2\\ \\ than\ the\ value\ of\ tan^{5} \theta +cot^{10} \theta \ is \\\ \\
(a) 1 \\\ \\
(b) 2 \\\ \\
(c) 3 \\\ \\
(d) 4 \\ \\
Read Solution
\Longrightarrow tan\theta +cot\theta =2\\\ \\ (we\ know\ cot\theta =\frac{1}{tan\theta })\\\ \\ \Longrightarrow tan\theta +\frac{1}{tan\theta } =2\\\ \\ \Longrightarrow \frac{tan^{2} \theta +1}{tan\theta } =2 \\\ \\ \Longrightarrow tan^{2} \theta +1=2tan\theta \\\ \\ \Longrightarrow tan^{2} \theta +1-2tan\theta =0 \\\ \\ ( use\ formula\ ( a-b)^{2} =a^{2} -2ab+b^{2})\\\ \\ \Longrightarrow ( tan \theta -1)^{2} =0 \\\ \\ \Longrightarrow tan \theta -1=0 \\\ \\ \Longrightarrow tan\theta =1 \\\ \\ ( we\ know\ tan\theta =\frac{1}{cot\theta }) \\\ \\ \Longrightarrow \frac{1}{cot\theta } =1\\\ \\ \Longrightarrow cot\theta =1\\\ \\ \\\ \\ now\\ \\ =\ tan^{5} \theta +cot^{10} \theta \\\ \\ ( put\ the\ value\ tan\theta =1\ \&\ cot\theta =1)\\\ \\ =1^{5} +1^{10}\\\ \\ =1+1\\ =2 \\\ \\
Option (b) is the right answer
(08) if\ sin\theta +cosec\theta =2\\ \\ than\ the\ value\ of\ sin^{5} \theta +cosec^{5} \theta \\ \\ when\ 0\degree \leqslant \theta \leqslant 90\degree \\\ \\
(a) 1 \\\ \\
(b) 2 \\\ \\
(c) 3 \\\ \\
(d) 100 \\ \\
Read Solution
\Longrightarrow sin\theta +cosec\theta =2\\\ \\ ( we\ know\ cosec\theta =\frac{1}{sin\theta })\\\ \\ \Longrightarrow sin\theta +\frac{1}{sin\theta } =2 \\\ \\ \Longrightarrow \frac{sin^{2} \theta +1}{sin\theta } =2 \\\ \\ \Longrightarrow sin^{2} \theta +1=2\ sin\theta \\\ \\ \Longrightarrow sin^{2} \theta +1-2sin\theta =0 \\\ \\ ( use\ formula\ ( a-b)^{2} =a^{2} -2ab+b^{2})\\\ \\ \Longrightarrow ( sin \theta -1)^{2} =0 \\\ \\ \Longrightarrow sin\theta -1=0\\\ \\ \Longrightarrow sin\theta =1\\\ \\ ( we\ know\ sin\theta =\frac{1}{cosec\theta })\\\ \\ \Longrightarrow \frac{1}{cosec\theta } =1\\\ \\ \Longrightarrow cosec\theta =1\\\ \\ \\\ \\ now\\ \\ =sin^{5} \theta +cosec^{5} \theta \\\ \\ =1^{5} +1^{5}\\\ \\ =1+1\\\ \\ =2\\\ \\
option (b) is the right answer
(09) if\ sec^{2} \theta +tan^{2} \theta =7\\ \\ than\ the\ value\ of\ \theta \ when\ 0\degree \leqslant \theta \leqslant 90\degree \ is \\\ \\
(a) 60\degree \\\ \\
(b) 30\degree \\\ \\
(c) 0\degree \\\ \\
(d) 90\degree \\ \\
Read Solution
\Longrightarrow sec^{2} \theta +tan^{2} \theta =7 \\\ \\ ( we\ know\ sec^{2} \theta =1+tan^{2} \theta )\\\ \\ \Longrightarrow \left( 1+tan^{2} \theta \right) +tan^{2} \theta =7\\\ \\ \Longrightarrow 2tan^{2} \theta =7-1\\\ \\ \Longrightarrow 2tan^{2} \theta =7-1\\\ \\ \Longrightarrow tan^{2} \theta =\frac{6}{2}\\\ \\ \Longrightarrow tan^{2} \theta =3 \\\ \\ \Longrightarrow tan\theta =\sqrt{3}\\\ \\ ( we\ know\ tan60\degree =\sqrt{3})\\\ \\ \Longrightarrow tan\theta =tan60\degree \\\ \\ \Longrightarrow \theta =60\degree
Option (a) is the right answer
(10) if\ sin17\degree =\frac{x}{y} \\ \\ than\ the\ value\ of\ ( sec17\degree -sin73\degree ) \ is\\\ \\
(a)\frac{y^{2}}{x\sqrt{y^{2} -x^{2}}} \\\ \\
(b)\frac{x^{2}}{y\sqrt{y^{2} -x^{2}}} \\\ \\
(c)\frac{x^{2}}{y\sqrt{x^{2} -y^{2}}} \\\ \\
(d)\frac{y^{2}}{x\sqrt{x^{2} -y^{2}}} \\ \\
Read Solution
sin17\degree =\frac{x}{y} \ =\frac{P}{H}\\\ \\ H^{2} =P^{2} +B^{2}\\\ \\ \Longrightarrow y^{2} =x^{2} +B^{2}\\\ \\ \Longrightarrow B^{2} =y^{2} -x^{2}\\\ \\ \Longrightarrow B=\sqrt{y^{2} -x^{2}}\\\ \\ now\\ \\ =sec17\degree -sin73\degree \\\ \\ =sec17\degree -sin( 90\degree -17\degree )\\\ \\ [ we\ know\ sin( 90\degree -\theta ) =cos\theta ]\\\ \\ =sec17\degree -cos17\degree \\\ \\ =\frac{H}{B} -\frac{B}{H}\\\ \\ =\frac{y}{\sqrt{y^{2} -x^{2}}} -\frac{\sqrt{y^{2} -x^{2}}}{y}\\\ \\ =\frac{y^{2} -\left(\sqrt{y^{2} -x^{2}}\right)^{2}}{y\sqrt{y^{2} -x^{2}}}\\\ \\ =\frac{y^{2} -\left( y^{2} -x^{2}\right)}{y\sqrt{y^{2} -x^{2}}}\\\ \\ = \frac{y^{2} -y^{2} +x^{2}}{y\sqrt{y^{2} -x^{2}}}\\\ \\ =\frac{x^{2}}{y\sqrt{y^{2} -x^{2}}} \\\ \\
Option (b) is the right answer
