objective Ques (310 results)
SSC CGL 202041)If \(5 cos^2\theta+1 =3sin^2\theta\), where, \(0^0<\theta<90^0\), then what is the value of \(tan\theta + sec\theta \over cot\theta + cosec\theta\)?
\(3+2\sqrt3\over3\)
\(5 cos^2\theta+1 =3sin^2\theta\); ⇒\(5(1-sin^2\theta)+1 =3sin^2\theta\); ⇒\(sin\theta = {\sqrt3\over2} = sin60^0\); \(\theta = 60^0\);
So \(tan\theta + sec\theta \over cot\theta + cosec\theta\) = \({tan60^0 + sec60^0 \over cot60^0 + cosec60^0 }= {3+2\sqrt3\over 3}\)
SSC CGL 202042)If \({sinA+cosA\over cosA}={17\over12}\) then the value of \(1-cosA\over sinA\) is:
\(1\over5\)
\({sinA\over cosA}+{cosA\over cosA} ={17\over12}\); ⇒ \(tanA+1 ={17\over 12}\); ⇒ tanA = \(5\over12\); Calculate \({1-cosA\over sinA }={1\over5}\)
SSC CGL 202043)If \(3sec^2\theta+tan\theta=7\), \(0^0<\theta<90^0\), then the value of \(cosec2\theta+cos\theta\over sin2\theta+cos\theta\) is:
\(2+\sqrt2\over4\)
\(3sec^2\theta+tan\theta=7\); ⇒ \(3(1+tan^2\theta)+tan\theta=7\); Solving we get \(\theta =45^0\);
\(\because\) Expression = \(cosec2\theta+cos\theta\over sin2\theta+cos\theta\) = \(2+\sqrt2\over4\)
SSC CGL 202044)The value of \({1-2sin^2\theta.cos^2\theta\over sin^4\theta+cos^4\theta}-1\) is :
0
\({1-2sin^2\theta.cos^2\theta\over sin^4\theta+cos^4\theta}-1={1-2sin^2\theta.cos^2\theta\over (sin^2\theta+cos^2\theta)^2-2sin^2\theta.cos^2\theta}-1 = {1-2sin^2\theta.cos^2\theta\over 1-2sin^2\theta.cos^2\theta}-1=1-1 = 0\)
SSC CGL 202045)What is the value of \(sin30^0+cos30^0-tan45^0\)?
\(\sqrt3-1\over2\)
\(sin30^0+cos30^0-tan45^0={1\over2}+{\sqrt3\over2}-1 ={\sqrt3-1\over2}\)
46)In the given figure, \(cos\theta\) is equal to:
SSC CGL 2020
\(5\over13\)
\(cos\theta={PR\over PQ}\); \(PR=\sqrt{PQ^2-QR^2} =\sqrt{13^2-12^2}=5\); \(\therefore cos\theta={5\over13}\)
SSC CGL 202047)The value of \(4[{(1-secA)^2+(1+secA)^2\over1+sec^2A}]\) is :
8
\(4[{(1-secA)^2+(1+secA)^2\over1+sec^2A}]=4[{(1+sec^2A-2secA+1+sec^2A+2secA)\over1+sec^2A}]=4[{(2+2sec^2A)\over1+sec^2A}]= 4\times2[{(1+sec^2A)\over1+sec^2A}]= 8\)
SSC CGL 202048)if 0 < A, B <\( 45^\circ\), \(\cos(A + B) = \frac{24}{25}\) and \(\sin(A - B) = \frac{15}{17}\), then \(\tan 2A\) is:
\(416\over87\)
tan 2A = tan((A + B) + (A - B)) =\(\frac{tan(A + B) + tan(A + B)}{1 - tan(A + B)tan(A + B)}\) ---(1);
\((\because tan(a + b) = \frac{tana + tanb}{1 - tana.tanb})
\); \(tan(A + B) = \frac{sin(A + B)}{cos(A - B)}\); ⇒
\(tan(A + B) = \frac{\sqrt{1 - cos^2(A + B)}}{cos(A + B)}\);⇒
\(tan(A + B) = \frac{\sqrt{49/25}}{(24/25)} = 7/24
\); \( tan(A - B) = \frac{sin(A - B)}{cos(A - B)}\);⇒
\(tan(A - B)= \frac{sin(A - B)}{\sqrt{1 - sin^2(A - B)}}\);⇒ \( tan(A - B) = \frac{15/17}{\sqrt{1 - (15/17)^2}}\);⇒
\(tan(A - B) = \frac{15/17}{\sqrt{64/17)^2}} = 15/8\);
From eq(1),
\(=\frac{\frac{7}{24} +\frac{15}{8}}{1 - \frac{7}{24}.\frac{15}{8}}
=\frac{416}{87}\)
SSC CGL 202049)If A lies in third quadrant, and 20 tan A = 21, then the value of \(\frac{5 \sin A - 2 \cos A}{4 \cos A - \frac{5}{7} \sin A}\) is:
1
20 tan A = 21; ⇒tan A = 21/20;
\(\frac{5 \sin A - 2 \cos A}{4 \cos A - \frac{5}{7} \sin A}
=\frac{\cos A(5 \frac{\sin A}{\cos A} - 2)}{\cos A(4 - \frac{5\sin A}{7\cos A})}
=\frac{5\tan A - 2}{4 - \frac{5}{7} \tan A}\);
On put the value of tan A,
\(= \frac{5 \times \frac{21}{20}- 2}{4 - \frac{5}{7} \times \frac{21}{20}}
= 1\)
SSC CGL 202050)If \((cos^2\theta-1)(1+tan^2\theta)+2tan^2\theta = 1\), \(0^0\leq\theta\leq90^0\) then \(\theta\) is :
\(45^0\)
\((cos^2\theta-1)(1+tan^2\theta)+2tan^2\theta = 1\) ; ⇒ \(-sin^2\theta.sec^2\theta+2 tan^2\theta=1\); ⇒ \({-sin^2\theta\over cos^2\theta}+2 tan^2\theta=1\); ⇒ \(-tan^2\theta+2 tan^2\theta=1\); \(tan^2\theta=1=tan^245^0\); \(\theta = 45^0\)