Class 12 NCERT Solutions- Mathematics Part I – Chapter 2 Inverse Trigonometric Functions – Miscellaneous Exercise on Chapter 2 | Set 1
Question 1. Find the value of
Solution:
We know that
Here,
Now, can be written as :
, where
Hence, the value of = π/6
Question 2. Find the value of
Solution:
We know that
Here,
Now, can be written as:
where
Hence, the value of = π/6
Question 3. Prove
Solution:
Let -(1)
sin x = 3/5
So,= 4/5
tan x = 3/4
Hence,
Now put the value of x from eq(1), we get
Now, we have
L.H.S
= –
Hence, proved.
Question 4. Prove
Solution:
Let
Then sin x = 8/17
cos x = = 15/17
Therefore,
-(1)
Now, let
Then, sin y = 3/5
= 4/5
-(2)
Now, we have:
L.H.S.
From equation(1) and (2), we get
=
=
= –
=
Hence proved
Question 5. Prove
Solution:
Let
Then, cos x = 4/5
= 3/5
-(1)
Now let
Then, cos y = 3/4
-(2)
Let
Then, cos z = 33/65
sin z = 56/65
-(3)
Now, we will prove that :
L.H.S.
From equation (1) and equation (2)
=
= –
=
=
Using equation(3)
=
Hence proved
Question 6. Prove
Solution:
Let
Then, sin x = 3/5
= 4/5
-(1)
Now, let
Then, cos y = 12/13 and sin y = 5/13
-(2)
Let
Then, sin z = 56/65 and cos z = 33/65
-(3)
Now, we have:
L.H.S.=
From equation(1) and equation(2)
=
= –
=
=
From equation (3)
=
Hence proved
Question 7. Prove
Solution:
Let
Then, sin x = 5/13 and cos x = 12/13.
-(1)
Let
Then, cos y = 3/5 and sin y = 4/5
-(2)
From equation(1) and (2), we have
R.H.S.
=
= –
=
=
L.H.S = R.H.S
Hence proved
Question 8. Prove
Solution:
L.H.S.
= –
=
=
=
=
=
=
= π/4
L.H.S = R.H.S
Hence proved
Question 9. Prove
Solution:
Let x = tan2θ
Then,
Now, we have
R.H.S =
L.H.S = R.H.S
Hence proved
Question 10. Prove
Solution:
Consider
By rationalizing
=
=
=
=
L.H.S = = x/2
L.H.S = R.H.S
Hence proved