Question

Find the value of tan (sin⁻¹$\frac{3}{5}$ + cos⁻¹$\frac{5}{\sqrt{34}}$).

Answer 1

Solution :

i ) Let sin^-1(3/5) = A

=> sinA = 3/5 ---( 1 )

ii ) cos²A

= 1 - sin²A

= 1 - ( 3/5 )²

= 1 - 9/25

= ( 25 - 9 )/25

= 16/25

cosA = 4/5 ---( 2 )

iii ) tanA = 3/4 --- ( 3 )

iv ) cos^-1(5/√34) = B

=> cosB = 5/√34 ----( 4 )

v ) sin² B

= 1 - cos²B

= 1 - ( 5/√34 )²

= 1 - 25/34

= ( 34 - 25 )/34

= 9/34

sinB = 3/√34 ---( 5 )

vi ) tanB = 3/5 --- ( 6 )

Now ,

tan{ sin^-1(3/5) + sin^-1(5/13) }

= tan( A + B )

= [ tanA + tanB ]/( 1 - tanAtanB )

= [ 3/4 + 3/5]/[ 1 - (3/4)(3/5) ]

= [ (15+12)/20 ]/[ (20-9)/20]

= (27/20)/(11/20)

= 27/11

••••

Answer 2

HELLO DEAR,



GIVEN:- tan (sin⁻¹$\frac{3}{5}$ + cos⁻¹$\frac{5}{\sqrt{34}}$).


let sin-¹ 3/5 = A

=> sinA = 3/5

cosA = √{1 - 9/25}

=> cosA = 4/5

tanA = sinA/cosA = 3/4

=> A = tan-¹3/4


similarly, cos-¹ 5/√34 = B

=> cosB = 5/√34

so, sinB = √{1 - 25/34}

=> sinB = 3/√34

tanB = sinB/cosB = 3/5

B = tan-¹3/5


now, tan (sin⁻¹$\frac{3}{5}$ + cos⁻¹$\frac{5}{\sqrt{34}}$). = tan(A + B)

=> tan(tan-¹3/4 + tan-¹ 3/5)

=> tan{tan-¹ (3/4 + 3/5)/(1 - 9/20)}

=> tan{tan-¹ (27/11)}

=> 27/11


I HOPE IT'S HELP YOU DEAR,
THANKS