NCERT Solutions
Class 11 Maths
Trigonometric Functions

Ex.Misc Q.10
Find sin , cos
and tan
for sin x =
, x in quadrant II
Here, x is in quadrant II.
i.e., < x < Π
=> <
<
Therefore, sin , cos
and tan
are all positive.
Given, sin x =
cos2 x = 1 – sin2 x
= 1 – ( )2
= 1 –
=
cos x = -√15 ÷ 4
[since cos x is negative in quadrant II]
Now, sin2 = (1 – cos x) ÷ 2
= {1 – (-√15 ÷ 4)} ÷ 2
= (1 + (√15 ÷ 4)) ÷ 2
= (4 + √15) ÷ 8
=> sin = √ [(4 + √15) ÷ 8]
[Since sin is positive]
=> sin = √ [{(4 + √15) ÷ 8} × (
)]
=> sin = √ [(8 + 2√15) ÷ 16]
=> sin = √ {(8 + 2√15)} ÷ 4
Again, cos2 = (1 + cos x) ÷ 2
= {1 + (-√15 ÷ 4)} ÷ 2
= (1 – (√15 ÷ 4)) ÷ 2
= (4 - √15) ÷ 8
=> cos = √ [(4 - √15) ÷ 8]
[Since cos is positive]
=> cos = √ [{(4 - √15) ÷ 8} × (
)]
=> cos = √ [(8 - 2√15) ÷ 16]
=> cos = √ {(8 - 2√15)} ÷ 4
Now, tan = (sin
) ÷ (cos
)
= [√ {(8 + 2√15)} ÷ 4] ÷ [ √ {(8 - 2√15)} ÷ 4]
= [√ (8 + 2√15)] ÷ [ √ (8 - 2√15)]
= √ [{(8 + 2√15)] ÷ (8 - 2√15)} × {(8 + 2√15)] ÷ (8 + 2√15)}]
= √ [(8 + 2√15)2] ÷ (64 - 60)]
= (8 + 2√15) ÷ 2
= 4 + √15
Therefore, value of sin , cos
and tan
are: √ {(8 + 2√15)} ÷ 4, √ {(8 - 2√15)} ÷ 4 and 4 + √15