Question

Prove that sin inverse 8/17+sin inverse 3/5= tan inverse 77/36

Answer 1

Answer:

Step-by-step explanation:

Let, a=sin^-1 8/17

∴ sin a=8/17         → (i)

and b=sin^-1 3/5

∴ sin b= 3/5           → (ii)

Now, cos^2 a= 1-sin^ a

cos a= √(1-sin^2 a)

        =√{1-(8/17)^2}

        =√(1-64/289)

        =15/17

∴ cos a=15/17

Again, cos b =√(1-sin^2b)

                     =√{1-(3/5)^2}

                     =√(1-9/25)

∴ cos b= 4/5

Let, tan a= sin a/ cos a

               =(8/17)/(15/17)

    ∴∴tan a= 8/15

Again let, tan b= sin b/cos b

                        =(3/5)/(4/5)

       ∴ tan b= 3/4

Now, tan(a+b)=(tan a + tan b)/(1 - tan a tan b)

                      = (8/15 + 3/4)/(1 - 8/15×3/4)

         ∴ tan(a+b)= 77/36

Hence, tan(a+b)=77/36

   ∴ a+b=tan^-1 77/36

Putting value of a and b we get,

sin^-1 8/17 + sin^-1 3/5= tan^-1 77/36

Hence proved.

Answer 2

Question:—

• Prove that sin inverse 8/17+sin inverse 3/5= tan inverse 77/36.

$\rule{200}2$

AnswEr:—

let sin-¹ 8/17 = x......(1)

→sin x = 8/17

cos x =√1-sin²x = √225/289 = 15/17

Then,

→tan x = sinx/cosx

→ tanx = 8/15

→ x = tan-¹ 8/15

comparing from eq(1) we get,

sin-¹ 8/17 = tan-¹ 8/15......(ii)

Similarly

let sin-¹3/5 = y .......(iii)

→ sin y= 3/5

cos y= √1-9/25 = 4/5

then ,

tan y = siny/cosy

→ tan y = 3/4

→ tan-¹ 3/4 = y

comparing from eq(iii) we get

→ sin-¹ 3/5 = tan-¹ 3/4 .....(iv)

putting the values of( ii) and (iv) into LHS

→tan-¹8/15 + tan-¹3/4

using formula tan-¹x + tan-¹ y = tan-¹(x+y/1-xy)

= tan-¹( 8/15+3/4/1-8/15×3/4)

= tan-¹ (77/36)

which is RHS

Hence proved