SSC CGL 202051)\({sin30^0sin60^0\over cos60^0cos30^0}-tan45^0\) का मान ज्ञात कीजिये |
0
\({{sin30^0sin60^0\over cos60^0cos30^0}-tan45^0}= {{1\over2}\times{\sqrt3\over2}\over{1\over2}\times{\sqrt3\over2}}-1 = 0\)
SSC CGL 202052)निम्न को हल कीजिए :
\(sin0^0\space sin30^0\space sin 45^0\space sin60^0\space sin90^0=?\)
0
\(sin0^0\space sin30^0\space sin 45^0\space sin60^0\space sin90^0=0\times{1\over2}\times{1\over\sqrt2}\times{\sqrt3\over2}\times1=0\)
SSC CGL 202053)यदि \((2\space sin A+cosecA)=2\sqrt2\), 0° < A < 90° है, तो \(2(sin^4A+cos^4A)\) का मान बताइए ।
1
\((2\space sin A+cosecA)=2\sqrt2\) ; ⇒ \((2\space sin A+{1\over sinA})=2\sqrt2\); ⇒ \(2sin^2A-2\sqrt2\space sinA+1=0\); ⇒ \(sinA={1\over\sqrt2}=sin45^0\); ⇒ \(A=45^0\); \(\therefore 2(sin^4A+cos^4A)= 2(sin^445^0+cos^445^0)=2({1\over4}+{1\over4})=2\times{1\over2}=1\)
SSC CGL 202054)निम्न को हल करें |
\({sin40^0\over cos50^0}+{cosec50^0\over sec40^0}-4cos50^0cosec40^0\)
-2
\({sin40^0\over cos50^0}+{cosec50^0\over sec40^0}-4cos50^0cosec40^0= {sin(90^0-50^0)\over cos 50^0}+{cosec(90^0-40^0)\over sec40^0}-4cos(90^0-40^0)cosec40^0\)= \({cos 50^0\over cos50^0}+{sec40^0\over sec40^0}-4sin40^0cosec40^0= 1+1-4 = -2\) \([\because sin\theta.cos\theta=1]\)
SSC CGL 202055)यदि x cosA - y sinA = 1 और x sinA + y cosA = 4 है, तो \(17x^2+17y^2\) का मान बताइए ।
289
Assume A = \(0^0\); \(\therefore x cos0^0-ysin0^0=1\); ⇒ x = 1; \(x sin0^0+y cos0^0=4\); y = 4; then \(17x^2+17y^2= 17+17\times(4)^2=289\)
SSC CGL 202056)\(\sqrt {tan^2 60^0+sin90^0}-2 \space tan 45^0\) का मान ज्ञात कीजिए ।
0
\(\sqrt {tan^2 60^0+sin90^0}-2 \space tan 45^0=\sqrt{(\sqrt{3})^2+1}-2\times1=0\)
SSC CGL 202057)यदि \(tan\space \theta-cot\space \theta = cosec\space \theta\), \(0^0<\theta<90^0\) है, तो \({2tan\space \theta-cos\space \theta\over \sqrt3cot\space \theta+sec\space \theta}\) का मान क्या होगा ?
\(4\sqrt3-1\over6\)
Click to Watch Video SolutionConvert all the trigonometric function in sin & cos.
\(\therefore {sin\theta\over cos\theta}-{cos\theta\over sin\theta }={1\over sin\theta}\) ; ⇒ \(sin^2\theta- cos^2\theta= cos\theta\); ⇒ \(2\space cos^2\theta+cos\theta-1=0\);
Solving we get \(\theta = 60^0\);
Put \(\theta = 60^0\) in given equation we get \(4\sqrt3-1\over6\)
SSC CGL 202058)\(cos\space 0^0\space cos\space 30^0\space cos \space 45^0\space cos\space 60^0\space cos\space 90^0\) का मान ज्ञात कीजिए ।
0
\(cos\space 0^0\space cos\space 30^0\space cos \space 45^0\space cos\space 60^0\space cos\space 90^0=1\times {\sqrt3\over2}\times{1\over\sqrt2}\times{1\over2}\times0\) = 0