, Avg: 01:40 , 50%✓ SSC CGL 201911)If \(sec\theta + tan\theta =p\), (p >1) then \({cosec\theta +1\over cosec\theta-1 } = ?\)
\(p^2\)
\(sec\theta+tan\theta=p\)________(1) \(sec\theta-tan\theta={1\over p}\)_______(2); from eq (1) and (2),\(sec\theta={p^2+1\over2p}\); \(cos\theta= {2p\over p^2+1}\); \(sin\theta={p^2-1\over p^2+1}\); Now calculate \(cosec\theta+1\over cosec\theta-1\) it will come out to be \(p^2\).
, Avg: 01:07 , 100%✓ SSC CGL 201912)The value of \(cosec(67^0 +\theta)-\)\(sec(23^0- \theta) +\)\(cos15^0cos35^0cosec55^0 \)\(cos60^0cosec75^0\) is :
\(1\over2\)
\((67^0 +\theta)-sec(23^0- \theta) \)\(+cos15^0cos35^0cosec55^0cos60^0cosec75^0\)= \((67^0 +\theta)-cosec(67^0+\theta) +\)\(cos15^0cos35^0sec35^0cos60^0sec15^0 \)\(= cos60^0={1\over2}\).
, Avg: 02:30 , 80%✓ SSC CGL 201913)If \(2cos^2\theta+3sin\theta=3\), where \(O^0<\theta<90^0\), then what is the value of \(sin^22\theta+cos^2\theta +\)\(tan^22\theta +cosec^22\theta\) ?
\(35\over6\)
, Avg: 02:07 , 60%✓ SSC CGL 201914)The value of \((1+cot\theta-cosec\theta) (1+cos\theta+sin\theta)\)\(sec\theta = ?\)
2
, Avg: 00:37 , 100%✓ SSC CGL 201915)The value of \({sec^2\theta\over cosec^2\theta}+\)\( {cosec^2\theta\over sec^2\theta}-\)\((sec^2\theta+cosec^2\theta)\) is :
-2
, Avg: 00:30 , 100%✓ SSC CGL 201916)The value of \({2(sin^6\theta+cos^6\theta)-3(sin^4\theta+cos^4\theta)}\over{cos^4\theta-sin^4\theta-2cos^2\theta}\) is :
1
, Avg: 00:14 , 100%✓ SSC CGL 201917)The value of \(sin^264^0+\)\(cos64^0sin26^0+\)\(2cos43^0cosec47^0\) is :
3
, Avg: 01:40 , 33%✓ SSC CGL 201918)The value of \({(sin\theta-cos\theta)(1+tan\theta+cot\theta)\over1+sin\theta cos\theta} = ?\)
\(sec\theta-cosec\theta\)
\({(sin\theta-cos\theta)(1+{sin\theta \over cos\theta}+ {cos\theta \over sin\theta})\over1+sin\theta cos\theta}\)
\((sinθ-cosθ)(sinθcosθ + sin^2θ +cos^2θ) \over (1+sinθcosθ)(sinθcosθ)\)
\(sec\theta-cosec\theta\)
, Avg: 00:49 , 100%✓ SSC CGL 201919)The value of \({sin\theta+cos\theta-1\over sin\theta-cos\theta+1}\times\)\({ tan^2\theta(cosec^2\theta-1)\over sec\theta-tan\theta}\) is :
, Avg: 00:58 , 42%✓ SSC CGL 201920)If \({sin\theta\over1+cos\theta}+\)\({1+cos\theta\over sin\theta}\)\(={4\over \sqrt3}\), \(0^0<\theta<90^0\), then the value of \((tan\theta+sec\theta)^{-1}\) is :
\(2-\sqrt3\)
\({{sin^2θ+1+cos^2θ+2cosθ} \over sinθ(1+cosθ)}={4 \over √ 3}\)
⇒ \({sinθ = }{√ 3 \over 2}\), θ = 60°