UPSC NDA Math Model Paper 7

Given that: sin−1(

3

5

)+tan−1(

1

7

)
Let sin−1(

3

5

)=θ
⇒sin(θ)=

3

5

As we know that, sin θ = (Opposite side / Hypotenuse)
From Pythagoras’ theorem,
(Hypotenuse)‌2 = (Adjacent side)‌2 + (Opposite side)‌2
⇒ (5)‌2 = (Adjacent side)‌2 + (3)‌2
⇒ Adjacent side = 4
As we know that, tan θ = (Opposite side / Adjacent side)
⇒ tan θ = (3/4)
⇒θ=tan−1(

3

4

)
We had already taken that, sin−1(

3

5

)=θ
So, sin−1(

3

5

)=θ=tan−1(

3

4

)
The given question can be written as: tan−1(

3

4

)+tan−1(

1

7

)
As we know that, tan−1A+tan−1B=tan−1{

A+B

1−(AB)

}
⇒tan−1{

3 4 + 1 7

1−( 3 4 × 1 7 )

}
⇒tan−1{

25

21

1−(

3

28

)}
As we know that, tan(π∕4)=1
⇒tan−1{

25 28

25 28

}=tan−1(1)=

π

4