SSC CGL 20191)यदि sin\({\theta}\) = \( { \sqrt3}\) cos\({\theta}\), 0°< θ < 90º है, तो (\(2sin^2{\theta}+sec^2{\theta}\)\(+sin{\theta}sec{\theta}+cosec{\theta}\)) का मान है ?
\( {33+10\sqrt3 \over 6}\)
\(\sin \theta = \sqrt{3} \cos \theta, 0^\circ < \theta < 90^\circ \Rightarrow \frac{\sin \theta}{\cos \theta} = \sqrt{3} \Rightarrow \tan \theta = \sqrt{3} \Rightarrow \theta = 60 \),
put 60 degree in the question and the answer will come out to be A
SSC CGL 20192)व्यंजक (\({cos^6\theta+sin^6\theta-1}\)) (\({tan^2\theta+cot^2\theta+2}\)) निम्न में से किसके बराबर है ?
-3
\((\cos^6 \theta + \sin^6 \theta - 1)(\tan^2 \theta + \cot^2 \theta + 2);\)
Assume any value of \theta,
Let the \( \theta\) be \(45^0\)
= \((\cos^6 45^0 + \sin^6 45^0- 1)(\tan^2 45^0 + \cot^2 45^0 + 2) = ((\frac{1}{\sqrt{2}})^6 + (\frac{1}{\sqrt{2}})^6 - 1)(1 + 1 + 2) = ((\frac{1}{8}) + (\frac{1}{8}) - 1)(4) = \frac{3}{4} \times 4 = -3\)
SSC CGL 20193)\({(2sinA)(1+sinA)}\over{1+sinA+cosA}\) का मान निम्न में से किसके बराबर है ?
1 + sin A - cos A
\(\frac{(2 \sin A)(1 + \sin A)}{1 + \sin A + \cos A}; = \frac{(2 \sin A + 2\sin^2 A)}{1 + \sin A + \cos A} = \frac{(2 \sin A + 2 - 2\cos^2 A)}{1 + \sin A + \cos A}\)
\((\because \sin^2 A + \cos^2 A = 1);\)
=\( \frac{(2 \sin A + 1 + \sin^2 A + \cos^2 A - 2\cos^2 A)}{1 + \sin A + \cos A} = \frac{((\sin A + 1)^2 - \cos^2 A)}{1 + \sin A + \cos A}\)
\((\because (a)^2 - (b)^2 = (a + b)(a - b));\)
=\( \frac{(1 + \sin A + \cos A)(1 + \sin A - \cos A)}{1 + \sin A + \cos A} = (1 + \sin A - \cos A)\)
SSC CGL 20194)\({(cos9^0 + sin81^0)(sec9^0 + cosec81^0)} \over {sin56^0sec34^0 + cos25^0cosec65^0}\) का मान है |
2
\(\frac{(\cos 9^\circ + \sin 81^\circ)(\sec 9^\circ + \ cosec 81^\circ)}{\sin 56^\circ sec 34^\circ + \cos 25^\circ \ cosec 65^\circ} = \frac{(\cos 9^\circ + \sin(90 - 9))(\sec 9^\circ + \ cosec(90 - 9))}{\sin 56^\circ sec(90 - 56) + \cos 25^\circ \ cosec(90 - 25)} = \frac{(\cos 9^\circ + \cos 9^\circ)(\sec 9^\circ + \sec 9^\circ)}{\sin 56^\circ cosec 56^\circ + \cos 25^\circ \sec 25^\circ} = \frac{(2\cos 9^\circ)(2 \sec 9^\circ)}{1 + 1} = \frac{4}{2} = 2\)
SSC CGL 20195)यदि \(\theta\) प्रथम चतुर्थांश (quadrant) में स्थित है और (\({(cos^2\theta-sin^2\theta)}={1\over 2}\) है, तो \({tan^22\theta+sin^23\theta}\) का मान है :
4
\(\cos^2 \theta - \sin^2 \theta = \frac{1}{2}\);
\(\cos^2 \theta - (1 - \cos^2 \theta) = \frac{1}{2};\)
\(2\cos^2 \theta - 1 = \frac{1}{2}; \) \( 2\cos^2 \theta = \frac{3}{2};\)
\(\cos^2 \theta = \frac{3}{4};\)
\(\cos \theta = \frac{\sqrt{3}}{2};\)
\(\theta = 30^0;\)
Now,
\(\tan^2 2\theta + \sin^2 3\theta; = \tan^2 2\times30^0 + \sin^2 3\times30^0 = \tan^2 60^0+ \sin^2 90^0 = 3 + 1 = 4\)
SSC CGL 20196)\(cosec(65^0+\theta)\)\(-sec(25^0-\theta)+\)\(tan^220^0-\)\(cosec^270^0\) का मान क्या है ?
-1
\(\ cosec(65^\circ + \theta) - \sec(25^\circ - \theta) + \tan^2 20^\circ - \ cosec^2 70^\circ =\ cosec(65^\circ + \theta) - \sec(90 - (65^\circ + \theta)) + \tan^2 20^\circ - \ cosec^2 (90 - 20)\)
\(=\ cosec(65^\circ + \theta) - \ cosec(65^\circ + \theta) + \tan^2 20^\circ - \sec^2 20^\circ = \tan^2 20^\circ - \sec^2 20^\circ = -(\sec^2 20^\circ -\tan^2 20^\circ ) = -1\)
SSC CGL 20197)\({(1+cos\theta)^2+sin^2\theta\over (cosec^2\theta-1)sin^2\theta}=?\)
\(2sec\theta(1+sec\theta)\)
\(\frac{\left(1+\cos\theta\right)^2+\sin^2\theta}{\left(\text{coec}^2\theta-1\right)\sin^2\theta}\)
\(=\frac{1 + \cos^2\theta + 2\cos\theta +\sin^2\theta}{1 -\sin^2\theta}
=\frac{2 + 2\cos\theta }{\cos^2\theta}
=2(\sec^2\theta+ \sec\theta)
=2\sec\theta(1 + \sec\theta)\)
SSC CGL 20198)\(({1-tan\theta\over 1- cot\theta})^2+1=?\)
\(sec^2\theta\)
\(\left(\frac{1-\tan\theta}{1-\cot\theta}\right)^2+1;\)
\((\frac{1-\tan\theta}{1-\frac{1}{tan\theta}})^2+1;\)
\((\frac{1-\tan\theta}{\frac{tan\theta - 1}{tan\theta}})^2+1;\)
\((-tan\theta)^2 + 1;\)
\(tan^2\theta + 1 = sec^2\theta\)
SSC CGL 20199)\(\sqrt{cot\theta+cos\theta\over cot\theta-cos\theta}\) निम्न में से किसके बराबर है ?
\(sec\theta+tan\theta\)
\(\sqrt{\frac{\cot\theta+\cos\theta}{\cot\theta-\cos\theta}}\)
\(= \sqrt{\frac{cos\theta(\frac{1}{sin\theta}+1)}{cos\theta(\frac{1}{sin\theta} - 1)}}
= \sqrt{\frac{1 + sin\theta}{1 - sin\theta} \times \frac{1 + sin\theta}{1 + sin\theta}}\)
\(= \sqrt{\frac{(1 + sin\theta)^2}{1 - sin^2\theta}}
= \frac{1 + sin\theta}{cos\theta}
= sec\theta + tan\theta\)
SSC CGL 201910)यदि 5 sinθ - 4 cosθ = 0, 0º < θ < 90º है, तो \({5 sin\theta- 2cos\theta \over 5sin\theta + 3cos\theta}\) का मान है |
\({2 \over 7}\)
\(5\sin\theta-4\cos\theta=0,0^{\circ}<\theta<90^{\circ}\)
\(= 5\sin\theta = 4\cos\theta\)
\(= \tan\theta = \frac{4}{5};\)
Now,
\(\frac{5\sin\theta-2\cos\theta}{5\sin\theta+3\cos\theta}
= \frac{\cos\theta(5\tan\theta-2)}{\cos\theta(5\tan\theta+3)}
= \frac{5\tan\theta-2}{5\tan\theta+3}
= \frac{5\times \frac{4}{5} -2}{5\times \frac{4}{5} +3}
= \frac{4 -2}{4 +3}
= \frac{2}{7}\)