Advertisements
Advertisements
प्रश्न
Find the value of `sin(tan^-1 (1/2) - cos^-1 (4/5))`
उत्तर
`sin(tan^-1 (1/2) - cos^-1 (4/5))`
a = `tan^-1 1/2`
tan a = `1/2 > 0` .......(Lies in I and III quadrant)
b = `cos^-1 (4/5)`
cos b = `4/5 > 0` .......(Lies in I and IV quadrant only)
`sin(tan^-1 (1/2) - cos^-1 (4/5)) = sin("a" - "b")`
= `sin "a" cos "b" - cos "a" sin "b"` .......[∵ sin(A – B) = sinA cosB – cosA sinB]
= `1/sqrt(5) (4/5) - 2/sqrt(5) * 3/5`
= `4/(5sqrt(5)) - 6/(5sqrt(5))`
= `(- 2)/(5sqrt(5)`
APPEARS IN
संबंधित प्रश्न
Find the domain of the following functions:
`tan^-1 (sqrt(9 - x^2))`
Find the domain of the following functions:
`1/2 tan^-1 (1 - x^2) - pi/4`
Find the value of `tan^-1(tan (5pi)/4)`
Find the value of `tan^-1 (tan(- pi/6))`
Find the value of `tan(tan^-1((7pi)/4))`
Find the value of `tan(tan^-1(1947))`
Find the value of `tan(tan^-1(- 0.2021))`
Find the value of `tan(cos^-1 (1/2) - sin^-1 (- 1/2))`
Find the value of `cos(sin^-1 (4/5) - tan^-1 (3/4))`
