Advertisements
Advertisements
प्रश्न
Prove the following result
`sin(cos^-1 3/5+sin^-1 5/13)=63/65`
उत्तर
LHS `=sin(cos^-1 3/5+sin^-1 5/13)`
`=sin[sin^-1sqrt(1-(3/5)^2)+sin^-1 5/13]`
`=sin[sin^-1 4/5+sin^-1 5/13]`
`=sin{sin^-1[4/5xxsqrt(1-(5/13)^2)+5/13xxsqrt(1-(4/5)^2)]}`
`=sin[sin^-1(48/65+15/65)]`
`=sin(sin^-1 63/65)`
`=63/65 =`RHS
APPEARS IN
संबंधित प्रश्न
Solve the equation for x:sin−1x+sin−1(1−x)=cos−1x
Find the principal values of the following:
`cos^-1(sin (4pi)/3)`
`sin^-1{(sin - (17pi)/8)}`
`sin^-1(sin12)`
Evaluate the following:
`cos^-1(cos3)`
Evaluate the following:
`cos^-1(cos4)`
Evaluate the following:
`tan^-1(tan (7pi)/6)`
Evaluate the following:
`sec^-1(sec (13pi)/4)`
Evaluate the following:
`cosec^-1(cosec (11pi)/6)`
Evaluate the following:
`cos(tan^-1 24/7)`
Solve: `cos(sin^-1x)=1/6`
Evaluate:
`cos(sec^-1x+\text(cosec)^-1x)`,|x|≥1
`5tan^-1x+3cot^-1x=2x`
Solve the following equation for x:
tan−1(x + 2) + tan−1(x − 2) = tan−1 `(8/79)`, x > 0
Solve the following equation for x:
`tan^-1((x-2)/(x-4))+tan^-1((x+2)/(x+4))=pi/4`
`sin^-1 5/13+cos^-1 3/5=tan^-1 63/16`
`tan^-1 2/3=1/2tan^-1 12/5`
Show that `2tan^-1x+sin^-1 (2x)/(1+x^2)` is constant for x ≥ 1, find that constant.
Find the value of the following:
`tan^-1{2cos(2sin^-1 1/2)}`
Solve the following equation for x:
`2tan^-1(sinx)=tan^-1(2sinx),x!=pi/2`
Write the difference between maximum and minimum values of sin−1 x for x ∈ [− 1, 1].
Write the value of sin (cot−1 x).
If tan−1 x + tan−1 y = `pi/4`, then write the value of x + y + xy.
Write the value of \[\sin^{- 1} \left( \frac{1}{3} \right) - \cos^{- 1} \left( - \frac{1}{3} \right)\]
Write the principal value of \[\tan^{- 1} 1 + \cos^{- 1} \left( - \frac{1}{2} \right)\]
Write the value of \[\sin^{- 1} \left( \sin\frac{3\pi}{5} \right)\]
Write the value of \[\cos^{- 1} \left( \cos\frac{14\pi}{3} \right)\]
If sin−1 x − cos−1 x = `pi/6` , then x =
sin\[\left[ \cot^{- 1} \left\{ \tan\left( \cos^{- 1} x \right) \right\} \right]\] is equal to
If α = \[\tan^{- 1} \left( \tan\frac{5\pi}{4} \right) \text{ and }\beta = \tan^{- 1} \left( - \tan\frac{2\pi}{3} \right)\] , then
If tan−1 (cot θ) = 2 θ, then θ =
The value of \[\sin\left( 2\left( \tan^{- 1} 0 . 75 \right) \right)\] is equal to
The domain of \[\cos^{- 1} \left( x^2 - 4 \right)\] is
Find : \[\int\frac{2 \cos x}{\left( 1 - \sin x \right) \left( 1 + \sin^2 x \right)}dx\] .
Prove that : \[\tan^{- 1} \left( \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} \right) = \frac{\pi}{4} + \frac{1}{2} \cos^{- 1} x^2 ; 1 < x < 1\].
Find the domain of `sec^(-1) x-tan^(-1)x`
Find the real solutions of the equation
`tan^-1 sqrt(x(x + 1)) + sin^-1 sqrt(x^2 + x + 1) = pi/2`
The value of sin `["cos"^-1 (7/25)]` is ____________.
