Advertisements
Advertisements
Question
Prove the following result:
`tan^-1 1/7+tan^-1 1/13=tan^-1 2/9`
Advertisements
Solution
LHS = `tan^-1 1/7+tan^-1 1/13`
`=tan^-1((1/7+1/13)/(1-1/7xx1/13))` `[becausetan^-1x+tan^-1y=tan^-1((x+y)/(1-xy))`
`=tan^-1((20/91)/(90/91))`
`=tan^-1 2/9=` RHS
APPEARS IN
RELATED QUESTIONS
If sin [cot−1 (x+1)] = cos(tan−1x), then find x.
If tan-1x+tan-1y=π/4,xy<1, then write the value of x+y+xy.
Find the principal values of the following:
`cos^-1(sin (4pi)/3)`
`sin^-1(sin (5pi)/6)`
`sin^-1(sin12)`
`sin^-1(sin2)`
Evaluate the following:
`cos^-1(cos3)`
Evaluate the following:
`tan^-1(tan (7pi)/6)`
Evaluate the following:
`sec^-1(sec (5pi)/4)`
Evaluate the following:
`sec^-1(sec (9pi)/5)`
Evaluate the following:
`\text(cosec)^-1(\text{cosec} pi/4)`
Evaluate the following:
`cosec^-1(cosec (11pi)/6)`
Evaluate the following:
`cot^-1{cot (-(8pi)/3)}`
Write the following in the simplest form:
`tan^-1(x/(a+sqrt(a^2-x^2))),-a<x<a`
Prove the following result
`tan(cos^-1 4/5+tan^-1 2/3)=17/6`
Evaluate:
`cot(sin^-1 3/4+sec^-1 4/3)`
Evaluate:
`sin(tan^-1x+tan^-1 1/x)` for x > 0
`sin^-1x=pi/6+cos^-1x`
`5tan^-1x+3cot^-1x=2x`
Solve the following equation for x:
`tan^-1 2x+tan^-1 3x = npi+(3pi)/4`
Solve the following equation for x:
tan−1(x −1) + tan−1x tan−1(x + 1) = tan−13x
Evaluate: `cos(sin^-1 3/5+sin^-1 5/13)`
`sin^-1 5/13+cos^-1 3/5=tan^-1 63/16`
`(9pi)/8-9/4sin^-1 1/3=9/4sin^-1 (2sqrt2)/3`
Evaluate the following:
`tan{2tan^-1 1/5-pi/4}`
Prove that
`tan^-1((1-x^2)/(2x))+cot^-1((1-x^2)/(2x))=pi/2`
Find the value of the following:
`cos(sec^-1x+\text(cosec)^-1x),` | x | ≥ 1
Solve the following equation for x:
`tan^-1((2x)/(1-x^2))+cot^-1((1-x^2)/(2x))=(2pi)/3,x>0`
If x < 0, then write the value of cos−1 `((1-x^2)/(1+x^2))` in terms of tan−1 x.
What is the value of cos−1 `(cos (2x)/3)+sin^-1(sin (2x)/3)?`
Write the value of cos−1 \[\left( \cos\frac{5\pi}{4} \right)\]
Write the value of \[\sin^{- 1} \left( \sin\frac{3\pi}{5} \right)\]
Wnte the value of the expression \[\tan\left( \frac{\sin^{- 1} x + \cos^{- 1} x}{2} \right), \text { when } x = \frac{\sqrt{3}}{2}\]
If α = \[\tan^{- 1} \left( \frac{\sqrt{3}x}{2y - x} \right), \beta = \tan^{- 1} \left( \frac{2x - y}{\sqrt{3}y} \right),\]
then α − β =
Let f (x) = `e^(cos^-1){sin(x+pi/3}.`
Then, f (8π/9) =
