SSC CGL 202021)The value of \({sin30^0sin60^0\over cos60^0cos30^0}-tan45^0\) is :
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 202022)Solve the following. \(sin0^0\space sin30^0\space sin 45^0\space sin60^0\space sin90^0=?\)
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 202023)If \((2\space sin A+cosecA)=2\sqrt2\), where 0° < A < 90°, then the value of \(2(sin^4A+cos^4A)\) is:
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 202024)Solve the following :
\({sin40^0\over cos50^0}+{cosec50^0\over sec40^0}-4cos50^0cosec40^0\)
-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 202025)If x cosA - y sinA = 1 and x sinA + y cosA = 4, then the value of \(17x^2+17y^2\) is:
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 202026)The value of \(\sqrt {tan^2 60^0+sin90^0}-2 \space tan 45^0\) is :
0
\(\sqrt {tan^2 60^0+sin90^0}-2 \space tan 45^0=\sqrt{(\sqrt{3})^2+1}-2\times1=0\)
SSC CGL 202027)If \(tan\space \theta-cot\space \theta = cosec\space \theta\), \(0^0<\theta<90^0\), then what is the value of \({2tan\space \theta-cos\space \theta\over \sqrt3cot\space \theta+sec\space \theta}\) ?
\(4\sqrt3-1\over6\)
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 202028)The value of \(cos\space 0^0\space cos\space 30^0\space cos \space 45^0\space cos\space 60^0\space cos\space 90^0\) is:
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
SSC CGL 202029)If \(5\space cos\theta-12\space sin\theta=0\),then what is the value of \({1+sin\space \theta +cos\space \theta\over 1-sin\space \theta+cos\space\theta}\) ?
\(3\over2\)
\(5 \cos \theta - 12 \sin \theta = 0\);
\(tan \theta = \frac{5}{12}\);
We know that \( tan \theta = \frac{perpendicular}{base}\) so,
By the triplet 5-12-13, Hypotenuse = 13 ;
\(sin \theta = \frac{5}{13}\); \(cos \theta = \frac{12}{13}\);
\(\frac{1 + \sin \theta + \cos \theta}{1 - \sin \theta + \cos \theta} = \frac{1 + \frac{5}{13} + \frac{12}{13}}{1 - \frac{5}{13} + \frac{12}{13}} = \frac{30}{20} = \frac{3}{2}\)
SSC CGL 202030)What is the value of \(\frac{cosec(78^\circ + \theta) - \sec(12^\circ - \theta) - \tan(67^\circ + \theta) + \cot(23^\circ - \theta)}{\tan 13^\circ \tan37^\circ \tan45^\circ \tan53^\circ \tan77^\circ}\) ?
0
\(\frac{cosec(78^\circ + \theta) - \sec(12^\circ - \theta) - \tan(67^\circ + \theta) + \cot(23^\circ - \theta)}{\tan 13^\circ \tan37^\circ \tan45^\circ \tan53^\circ \tan77^\circ}\) = \(-tan(67^0+\theta)+tan(67^0+\theta)\over(tan13^0.cot13^0)(tan37^0.cot37^0).tan45^0\) \([\because tan(90^0-\theta)=cot\theta]\)
= \(0-0\over1\times1\times1\) = 0