CBSE Class XI

Question 10:

Answer

Here, x is in quadrant II.

Question 9:

Answer

Here, x is in quadrant III.

It is given that cos x = −1/3.

Question 8:

tan x = −4/3, x in quadrant II

Answer

 

As x is in quadrant II, cosx is negative.

Question 7:

Answer

Question 6:

Answer

Question 5:

Prove that: sin x + sin 3x + sin 5x + sin 7x = 4 cos x cos 2x sin 4x

Answer

Question 4:

Answer

Question 3:

Answer

Question 2:

Prove that: (sin 3x + sin x) sin x + (cos 3x − cos x) cos x = 0

Answer

L.H.S.  = (sin 3x + sin x) sin x + (cos 3x − cos x) cos x = 0

=sin 3x sin x + sin² x + cos 3x cos x – cos² x

=cos 3x cos x + sin 3x sin x – (cos²x – sin² x)

= cos(3x – x) – cos 2x             [cos(A – B) = cos A cos B + sin A sin B]

=cos 2x – cos 2x

= 0

= R.H.S.

Question 1:

Answer

Question 9:

Find the general solution of the equation sin x + sin 3x + sin 5x = 0

Answer

sin x + sin 3x +  sin 5x = 0

(sin x + sin 5x) + sin 3x = 0

Question 8:

Find the general solution of the equation sec² 2x = 1 – tan 2x

Answer

sec²2x = 1- tan 2x

⇒ 1 + tan² 2x= 1 = 1 – tan 2x

⇒ tan² 2x + tan 2x = 0

⇒ tan 2x(tan 2x + 1) = 0

⇒tan 2x = 0                       or                    tan 2x + 1 = 0

Now, tan 2x = 0

⇒ tan 2x = tan 0

⇒ 2x = nπ + 0, where n ∈ Z

⇒ x = nπ/2, where n ∈ Z

tan 2x + 1 = 0

Question 7:

Find the general solution of the equation sin 2x + cos x = 0

Answer

Question 6:

Find the general solution of the equation cos 3x + cos x – cos 2x = 0

Answer

Question 5:

Find the general solution of the equation cos 4x = cos 2x

Answer

Question 4:

Find the general solution of cosec x = -2

Answer

Question 3:

Find the principal and general solutions of the equation cot x = −√3

Answer

Question 2:

Find the principal and general solutions of the equation sec x = 2

Answer

Therefore, the general solution is x = 2nπ ± π/3, where n ∈ Z

Question 1:

Find the principal and general solutions of the equation tan x = √3

Answer

Question 25:

Prove that: cos 6x = 32 cos⁶ x − 48 cos⁴ x + 18 cos² x – 1

Answer

L.H.S. = cos 6x

= cos 3(2x)

= 4 cos³ 2x − 3 cos2x [cos 3A = 4 cos³ A − 3 cosA]

= 4 [(2 cos² x − 1)³ − 3 (2 cos² x − 1) [cos 2x = 2 cos² x − 1]

= 4 [(2 cos² x)³ − (1)³ − 3 (2 cos² x)² + 3 (2 cos² x)] − 6cos² x + 3

= 4 [8cos⁶x − 1 − 12 cos⁴x + 6 cos²x] − 6 cos²x + 3

= 32 cos⁶x − 4 − 48 cos⁴x + 24 cos² x − 6 cos²x + 3

= 32 cos⁶x − 48 cos⁴x + 18 cos²x – 1

= R.H.S.