objective Ques (310 results)
SSC CGL 20201)यदि \(6tan\theta-5\sqrt3sec\theta+12cot\theta=0 \),\(0^0 <\theta< 90^0\) है, तो \(cosec\theta +sec\theta\) का मान ज्ञात कीजिए।
\({2(3+\sqrt3) \over 3}\)
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)दी गई आकृति, समकोण त्रिभुज में, \(cosec\theta\) का मान कितना है ?
SSC CGL 2020
13/5
\(cosec\theta = {H /P}\)
SSC CGL 20193)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 20194)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 20195)\({(2sinA)(1+sinA)}\over{1+sinA+cosA}\) 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 20196)\({(cos9^0 + sin81^0)(sec9^0 + cosec81^0)} \over {sin56^0sec34^0 + cos25^0cosec65^0}\) 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 20197)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 20198)\(cosec(65^0+\theta)\)\(-sec(25^0-\theta)+\)\(tan^220^0-\)\(cosec^270^0\) 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 20199)\({(1+cos\theta)^2+sin^2\theta\over (cosec^2\theta-1)sin^2\theta}=?\)
\(2sec\theta(1+sec\theta)\)
\(\frac{\left(1+\cos\theta\right)^2+\sin^2\theta}{\left(\text{coec}^2\theta-1\right)\sin^2\theta}\)
\(=\frac{1 + \cos^2\theta + 2\cos\theta +\sin^2\theta}{1 -\sin^2\theta}
=\frac{2 + 2\cos\theta }{\cos^2\theta}
=2(\sec^2\theta+ \sec\theta)
=2\sec\theta(1 + \sec\theta)\)
SSC CGL 201910)\(({1-tan\theta\over 1- cot\theta})^2+1=?\)
\(sec^2\theta\)
\(\left(\frac{1-\tan\theta}{1-\cot\theta}\right)^2+1;\)
\((\frac{1-\tan\theta}{1-\frac{1}{tan\theta}})^2+1;\)
\((\frac{1-\tan\theta}{\frac{tan\theta - 1}{tan\theta}})^2+1;\)
\((-tan\theta)^2 + 1;\)
\(tan^2\theta + 1 = sec^2\theta\)