SSC CGL 2020 objective Ques (34 results)
SSC CGL 20201)If \(6tan\theta-5\sqrt3sec\theta+12cot\theta=0\) where, \(0 <\theta< 90\), then value of \(cosec\theta +sec\theta\) is:
\({2(3+\sqrt3) \over 3}\)
since, \(0 <\theta< 90\), it is always advisable in competitive exams to use \(\theta = 30, 45 and 60\) In the instant question \(\theta = 60 \) will satisfy the condition \(6tan\theta-5\sqrt3sec\theta+12cot\theta=0\) , therefore put \(\theta = 60 \) in \(cosec\theta +sec\theta\) and calulate answer.
2)In a right triangle shown in the figure, what is the value of \(cosec\theta\) ?
SSC CGL 2020
13/5
\(cosec\theta = {H /P}\)
SSC CGL 20203)Solve the following : \({2sin22^0\over cos68^0}-\)\({2cot75^0\over 5 tan15^0}\)\(-{8tan45^0tan20^0tan40^0tan50^0tan70^0\over5}\)
0
\({2sin22^0\over cos68^0}-{2cot75^0\over 5 tan15^0}-{8tan45^0tan20^0tan40^0tan50^0tan70^0\over5}\) =\({2sin22^0\over cos(90-22)^0}-{2cot(90-15)^0\over 5 tan15^0}-{8tan45^0tan(90-70)^0tan(90-40)^0tan50^0tan70^0\over5}\) = \({2sin22^0\over sin22^0}-{2cot75^0\over 5 cot75^0}-{8tan20^0cot20^0tan40^0cot40^0\over5}\) = \(2-{2\over5}-{8\over5}\) = 0
SSC CGL 20204)If, \(5cot\theta=3\), find the value of \(6sin\theta-3cos\theta\over7sin\theta+3cos\theta\) .
\(21\over44\)
\(5cot\theta=3\); \(cot\theta={3\over5}\); (dividing numerator and denominator by \(sin\theta\)); \({6sin\theta-3cos\theta\over7sin\theta+3cos\theta}={{{6sin\theta\over sin\theta}-{3cos\theta\over sin \theta}}\over{{7sin\theta\over sin\theta}+{3cos\theta\over sin \theta}}}\)= \({6-3cot\theta\over7+3cot\theta}={6-3\times\frac{3}{5}\over7+3\times\frac{3}{5}}\) =\(21\over44\)
SSC CGL 20205)The value of \((cosecA+cotA+1)\)\((cosec A-cot A+1)\)\(-2 cosec A\) is:
2
(cosecA + cotA + 1)(cosecA − cotA + 1) − 2cosecA = \((cosecA + 1)^2 − (cotA)^2 − 2cosecA\); { \(a^2-b^2= (a+b)(a-b)\)}; = \(cosec^2A +1+2cosecA-cot^2A-2cosecA\)\((cosec^2A - cot^2A = 1)\) = 1 + 1 = 2
SSC CGL 20206)The value of (\(cos10^0 \) \(cos 30^0\) \(cos 50^0\) \(cos 70^0\) \(cos 90^0\)) is :
0
\(cos 10^0 cos 30^0 cos 50 ^0 cos 70^0 cos 90^0= cos10^0. cos30^0. cos50^0. cos70^0\times0 = 0\).
SSC CGL 20207)The value of \({3(1-2sin^2x)}\over{cos^2x-sin^2x}\) is :
3
\({{3(1-2sin^2x)}\over{cos^2x-sin^2x}}={ {3(1-2sin^2x)}\over{1-sin^2x-sin^2x}}\) =\({{3(1-2sin^2x)}\over(1-2sin^2x)}= 3\)
SSC CGL 20208)The value of \((cosec30^0-tan45^0)\)\(cot60^0tan30^0\) is :
\(1\over3\)
\((cosec30^0-tan45^0)cot60^0tan30^0 = (2-1){1\over \sqrt3}\times {1\over \sqrt3}={1\over3}\)
SSC CGL 20209)If \(cot\theta+tan\theta=2sec\theta\), where \(0^0<\theta<90^0\), then the value of \(tan2\theta-sec\theta\over cot2\theta+cosec\theta\) is :
\(2\sqrt3-1\over11\)
\(cot\theta+tan\theta=2sec\theta\); \({cos\theta\over sin\theta}+{sin\theta\over cos\theta}={2\over cos\theta}\); \({cos^2\theta+sin^2\theta\over sin\theta cos\theta}={2\over\cos\theta}\); \(sin\theta={1\over2} =sin30^0\); so put \(\theta=30^0\);
\({tan 60^0-sec30^0\over cot60^0+cosec30^0}= {2\sqrt{3}-1\over11}\)
SSC CGL 202010)Solve the following. \(({sin27^0\over cos63^0})-({cos27^0\over sin63^0})^2\)
0
\(({sin27^0\over cos63^0})-({cos27^0\over sin63^0})^2=[{sin27^0\over cos(90-27^0)}]-[{cos27^0\over sin(90-27^0)}]^2\) = \([{sin27^0\over sin27^0}]-[{cos27^0\over cos27^0}]^2 = 1 - (1)^2 =0\)