Class 11_Chapter 3_Lecture_2

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Class XI
Mathematics
Chapter- 3
Trigonometric Functions
Lecture - 2
Dr. Pranav Sharma
Maths Learning Centre. Jalandhar.
Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar

youtube.com/@MathematicsOnlineLectures
TRIGONOMETRIC (OR CIRCULAR) FUNCTIONS
Let 𝐗′
𝑶𝐗 and 𝒀𝑶𝒀′ be the coordinate axes. Taking 𝑶 as the centre and a unit radius,
draw a circle, cutting the coordinate axes at 𝑨, 𝑩, 𝑨′
and 𝑩′
, as shown in the figure.
Suppose that a moving point starts from 𝑨 and moves along the circumference of the
circle in an anticlockwise direction. Let it cover an arc length 𝜽 and take the final
position 𝑷(𝒙, 𝒚) .
Join 𝑶𝑷.
Then, ∠𝑨𝑶𝑷 = 𝜽
[⋅.⋅ ∠𝑨𝑶𝑷 =
𝐚𝐫𝐜𝑨𝑷
𝐫𝐚𝐝𝐢𝐮𝐬𝑶𝑷
=
𝜽
𝟏
= 𝜽𝒄
, using 𝜽 =
𝒍
𝒓
].

Now, the six circular functions may be defined as:
(i) 𝐜𝐨𝐬 𝜽 = 𝒙 (ii) 𝐬𝐢𝐧 𝜽 = 𝒚
(iii) 𝐬𝐞𝐜 𝜽 =
𝟏
𝒙
, 𝒙 ≠ 𝟎 (iv) 𝐜𝐨𝐬𝐞𝐜 𝜽 =
𝟏
𝒚
, 𝒚 ≠ 𝟎
(v) 𝐭𝐚𝐧 𝜽 =
𝒚
𝒙
, 𝒙 ≠ 𝟎 (vi) 𝐜𝐨𝐭 𝜽 =
𝒙
𝒚
, 𝐲 ≠ 𝟎
We have
(i) 𝐬𝐞𝐜 𝜽 =
𝟏
𝐜𝐨𝐬 𝜽
(ii) 𝐜𝐨𝐬𝐞𝐜 𝜽 =
𝟏
𝐬𝐢𝐧 𝜽
(iii) 𝐭𝐚𝐧 𝜽 =
𝐬𝒊𝒏𝜽
𝐜𝐨𝐬 𝜽
(iv) 𝐜𝐨𝐭 𝜽 =
𝟏
𝐭𝐚𝐧 𝜽
=
𝐜𝐨𝐬 𝜽
𝐬𝐢𝐧 𝜽

For any acute angle, say 𝜽, we have
(i) 𝐜𝐨𝐬𝟐
𝜽 + 𝐬𝐢𝐧𝟐
𝜽 = 𝟏 (ii) 𝟏 + 𝐭𝐚𝐧𝟐
𝜽 = 𝐬𝐞𝐜𝟐
𝜽 (iii) 𝟏 + 𝐜𝐨𝐭𝟐
𝜽 = 𝐜𝐨𝐬𝐞𝐜𝟐
𝜽
Let 𝐗′
𝑶𝑿 and 𝐘𝑶𝐘′
be the coordinate axes. Taking 𝑶 as the centre and a unit radius,
draw a circle, meeting 𝑶𝐗 at 𝑨.
Let 𝑷(𝒙, 𝒚) be a point on the circle with ∠𝑨𝑶𝑷 = 𝜽.
Join 𝑶𝑷. Draw 𝑷𝑴 ⊥ 𝑶𝑨.
Then, 𝒄𝒐𝒔𝜽 = 𝒙 and 𝐬𝐢𝐧 𝜽 = 𝒚.
(i) From right 𝚫𝑶𝑴𝑷, we have
𝑶𝑴𝟐
+ 𝑷𝑴𝟐
= 𝑶𝑷𝟐
⇒ 𝒙𝟐
+ 𝒚𝟐
= 𝟏
⇒ 𝐜𝐨𝐬𝟐
𝜽 + 𝐬𝐢𝐧𝟐
𝜽 = 𝟏…(i)
(ii) Dividing (i) by 𝒄𝒐𝒔𝟐
𝜽 we get,𝟏 +
𝒔𝒊𝒏𝟐𝜽
𝒄𝒐𝒔𝟐𝜽
=
𝟏
𝒄𝒐𝒔𝟐𝜽
⇒ 𝟏 + 𝐭𝐚𝐧𝟐
𝜽 = 𝐬𝐞𝐜𝟐
𝜽
(iii) Dividing both sides of (i) by 𝒔𝒊𝒏𝟐
𝜽 , we get
𝐜𝐨𝐬𝟐𝜽
𝐬𝐢𝐧𝟐𝜽
+ 𝟏 =
𝟏
𝐬𝐢𝐧𝟐𝜽
⇒ 𝐜𝐨𝐭𝟐
𝜽 + 𝟏 = 𝐜𝐨𝐬𝐞𝐜𝟐
𝜽 ⇒ 𝟏 + 𝐜𝐨𝐭𝟐
𝜽 = 𝐜𝐨𝐬𝐞𝐜𝟐
𝜽.

NEGATIVE ARC LENGTH If a point moves in a circle then the arc length covered by it
is said to be positive or negative depending on whether the point moves in the
anticlockwise or clockwise direction respectively.
For any acute angle, say 𝜽, we have
(i) 𝐜𝐨𝐬 (−𝜽) = 𝐜𝐨𝐬 𝜽 (ii) 𝐬𝐢𝐧 (−𝜽) = − 𝐬𝐢𝐧 𝜽 (iii) 𝐭𝐚𝐧 (−𝜽) = − 𝐭𝐚𝐧 𝜽
(iv) 𝐜𝐨𝐭 (−𝜽) = − 𝐜𝐨𝐭 𝛉 (v) 𝐬𝐞𝐜 (−𝜽) = 𝐬𝐞𝐜 𝛉 (vi) 𝐜𝐨𝐬𝐞𝐜 (−𝜽) = − 𝐜𝐨𝐬𝐞𝐜 𝜽
Let 𝐗′
𝑶𝐗 and 𝐘𝑶𝐘′
be the coordinate axes. With 𝑶 as the
centre, draw a circle of unit radius, meeting 𝑶𝑿 at 𝑨(𝟏, 𝟎) .
Let 𝑷(𝒙, 𝒚) be a point on the circle such that ∠𝑨𝑶𝑷 = 𝜽.
Let∠𝑨𝑶𝑸 = −𝜽. Then the coordinate of 𝑸 are 𝑸(𝒙, −𝒚). So,
(i) 𝒄𝒐𝒔(−𝜽) = 𝒙 = 𝒄𝒐𝒔𝜽
(ii) 𝒔𝒊𝒏(−𝜽) = −𝒚 = −𝒔𝒊𝒏(𝜽)

Signs of Trigonometric Functions in Various Quadrants
Quadrant Signs of various 𝐓‐functions
I All 𝐓‐functions are positive.
II 𝐬𝐢𝐧 𝜽 and 𝐜𝐨𝐬𝐞𝐜 𝜽 are positive.
All others are negative.
III 𝐭𝐚𝐧 𝜽 and 𝐜𝐨𝐭 𝜽 are positive.
IV 𝐜𝐨𝐬 𝜽 and 𝒔𝒆𝒄𝜽 are positive.

Values of 𝑻‐functions of Some Particular Angles
𝜽 → 𝟎 𝝅
𝟔
𝝅
𝟒
𝝅
𝟑
𝝅
𝟐
𝝅 𝟑𝝅
𝟐
𝟐𝝅
𝐬𝐢𝐧 𝛉 𝟎 𝟏
𝟐
𝟏
√𝟐
√𝟑
𝟐
𝟏 𝟎 −𝟏 𝟎
𝐜𝐨𝐬 𝛉 𝟏 √𝟑
𝟐
𝟏
√𝟐
𝟏
𝟐
𝟎 −𝟏 𝟎 𝟏
𝒕𝒂𝒏 𝛉 𝟎 𝟏
√𝟑
𝟏 √𝟑 not
defined
𝟎 not
defined
𝟎

Domain and Range of Trigonometric Functions
The domain and range of each one of the six trigonometric functions is given below.
𝐓‐function Domain Range
𝐬𝐢𝐧 𝒙 𝑹 [−𝟏, 𝟏]
𝐜𝐨𝐬 𝒙 𝑹 [−𝟏, 𝟏]
𝐭𝐚𝐧 𝒙
{𝒙 ∈ 𝑹 ∶ 𝒙 ≠
(𝟐𝐧 + 𝟏)𝝅
𝟐
, 𝒏 ∈ 𝑰}
𝑹
𝐜𝐨𝐭 𝒙 {𝒙 ∈ 𝑹: 𝒙 ≠ 𝒏𝝅, 𝒏 ∈ 𝑰} 𝑹
𝐜𝐨𝐬𝐞𝐜 𝒙 {𝒙 ∈ 𝑹: 𝒙 ≠ 𝒏𝝅, 𝒏 ∈ 𝑰} 𝑹−] − 𝟏, 𝟏[
𝐬𝐞𝐜 𝒙
{𝒙 ∈ 𝑹 ∶ 𝒙 ≠
(𝟐𝐧 + 𝟏)𝝅
𝟐
, 𝒏 ∈ 𝑰}
𝑹−] − 𝟏, 𝟏[

PERIODIC FUNCTIONS A function 𝒇(𝒙) is said to be periodic if there exists a constant
real number 𝒑 such that 𝒇(𝒙 + 𝒑) = 𝒇(𝒙) for all 𝒙.
The least positive value of 𝒑 for which 𝒇(𝒙 + 𝒑) = 𝒇(𝒙) is called the period of 𝒇(𝒙) .
AN IMPORTANT RESULT If 𝒇(𝒙) is a periodic function with period 𝒑 then 𝒇(𝒂𝒙 + 𝒃)
with 𝒂 > 𝟎 is a periodic function with period (
𝐛
𝐚
).

𝐬𝐢𝐧 (𝜽 + 𝟐𝝅) = 𝐬𝐢𝐧 𝜽 and 𝐜𝐨𝐬 (𝜽 + 𝟐𝝅) = 𝒄𝒐𝒔 𝛉.
Let 𝑿′𝑶𝑿 and 𝒀𝑶𝒀′ be the coordinate axes. With 𝑶 as the centre, draw a circle of unit
radius, meeting 𝑶𝐗 at 𝑨. The circumference of this circle of unit radius is 𝟐𝝅.
Let 𝑷 be a point on this circle such that arc 𝑨𝑷 = 𝜽.
If we start from the point 𝑨, move along the circle and after making a complete
revolution, reach 𝑷 then the length of the arc covered is
(𝜽 + 𝟐𝝅) .
Since the trigonometric functions are defined in terms of
the coordinates of 𝑷, we have 𝐜𝐨𝐬 (𝜽 + 𝟐𝝅) = 𝐜𝐨𝐬 𝜽 and
𝐬𝐢𝐧 (𝜽 + 𝟐𝝅) = 𝐬𝐢𝐧 𝜽.

(i) Since 𝟐𝝅 is the least angle for which 𝐜𝐨𝐬 (𝜽 + 𝟐𝝅) = 𝐜𝐨𝐬 𝜽 and 𝐬𝐢𝐧 (𝜽 + 𝟐𝝅) =
𝐬𝐢𝐧 𝜽, it follows that
𝐜𝐨𝐬 𝜽 and 𝐬𝐢𝐧 𝜽 are periodic functions, each with period 𝟐𝝅.
Similarly, 𝐬𝐞𝐜 𝜽 and 𝒄𝒐𝒔𝒆𝒄 𝜽 are periodic, each with period 𝟐𝝅; and
𝐭𝐚𝐧 𝜽 and 𝐜𝐨𝐭 𝜽 are periodic, each with period 𝝅.
A constant function is periodic, having no period.
(ii) For any real number 𝜽 and any integer 𝒏, we have:
𝐬𝐢𝐧 (𝛉 + 𝟐𝒏𝝅) = 𝐬𝐢𝐧 𝜽, 𝐜𝐨𝐬 (𝜽 + 𝟐𝒏𝝅) = 𝐜𝐨𝐬 𝜽
𝐭𝐚𝐧 (𝜽 + 𝒏𝝅) = 𝐭𝐚𝐧 𝜽, 𝐜𝐨𝐭 (𝜽 + 𝒏𝝅) = 𝐜𝐨𝐭 𝜽
𝐬𝐞𝐜 (𝜽 + 𝟐𝒏𝝅) = 𝐬𝐞𝐜 𝜽, 𝐜𝐨𝐬𝐞𝐜 (𝜽 + 𝟐𝒏𝝅) = 𝐜𝐨𝐬𝐞𝐜 𝜽

If 𝒔𝒆𝒄 𝜽 =
−𝟏𝟑
𝟏𝟐
and 𝜽 lies in the second quadrant, find the values of all the other five
trigonometric functions.
We know that in the second quadrant, 𝒔𝒊𝒏𝜽 and 𝒄𝒐𝒔𝒆𝒄𝜽 are positive and all the other
trigonometric functions are negative.
Now, 𝐬𝐞𝐜 𝜽 =
−𝟏𝟑
𝟏𝟐
⇒ 𝐜𝐨𝐬 𝜽 =
𝟏
𝐬𝐞𝐜 𝜽
=
−𝟏𝟐
𝟏𝟑
.
𝒕𝒂𝒏𝟐
𝜽 = (𝐬𝐞𝐜𝟐
𝜽 − 𝟏) = {(
−𝟏𝟑
𝟏𝟐
)
𝟐
− 𝟏} = (
𝟏𝟔𝟗
𝟏𝟒𝟒
− 𝟏) =
𝟐𝟓
𝟏𝟒𝟒
⇒ 𝐭𝐚𝐧 𝜽 = −√
𝟐𝟓
𝟏𝟒𝟒
=
−𝟓
𝟏𝟐
⇒ 𝐜𝐨𝐭 𝜽 =
𝟏
𝐭𝐚𝐧 𝜽
=
−𝟏𝟐
𝟓
.
Also, 𝐬𝐢𝐧 𝜽 = 𝐭𝐚𝐧 𝜽 ⋅ 𝐜𝐨𝐬 𝜽 = (
−𝟓
𝟏𝟐
) × (
−𝟏𝟐
𝟏𝟑
) =
𝟓
𝟏𝟑
. And, 𝒄𝒐𝒔𝒆𝒄 𝜽 =
𝟏
𝐬𝐢𝐧 𝜽
=
𝟏𝟑
𝟓
.
Hence, 𝒄𝒐𝒔 𝜽 =
−𝟏𝟐
𝟏𝟑
; 𝐭𝐚𝐧 𝜽 =
−𝟓
𝟏𝟐
; 𝐜𝐨𝐭 𝜽 =
−𝟏𝟐
𝟓
; 𝐬𝐢𝐧 𝜽 =
𝟓
𝟏𝟑
and 𝐜𝐨𝐬𝐞𝐜 𝜽 =
𝟏𝟑
𝟓
.

If 𝐬𝐢𝐧 𝜽 =
−𝟒
𝟓
and 𝝅 < 𝜽 <
𝟑𝝅
𝟐
, find the values of all the other five trigonometric
functions.
Clearly, 𝜽 lies in the third quadrant in which 𝐭𝐚𝐧 𝜽 and 𝒄𝒐𝒕𝜽 are positive and all the
other trigonometric functions are negative.
Now, 𝐬𝐢𝐧 𝜽 =
−𝟒
𝟓
⇒ 𝐜𝐨𝐬𝐞𝐜 𝜽 =
𝟏
𝐬𝐢𝐧 𝜽
=
−𝟓
𝟒
.
𝒄𝒐𝒕𝟐
𝟔 = (𝐜𝐨𝐬𝐞𝐜𝟐
𝜽 − 𝟏) = (
𝟐𝟓
𝟏𝟔
− 𝟏) =
𝟗
𝟏𝟔
⇒ 𝐜𝐨𝐭 𝜽 = +√
𝟗
𝟏𝟔
=
𝟑
𝟒
⇒ 𝐭𝐚𝐧 𝜽 =
𝟏
𝐜𝐨𝐭 𝜽
=
𝟒
𝟑
.
Also, 𝐜𝐨𝐬 𝜽 = 𝐜𝐨𝐭 𝜽 𝐬𝐢𝐧 𝜽 = (
𝟑
𝟒
) × (
−𝟒
𝟓
) =
−𝟑
𝟓
. 𝒔𝒆𝒄 𝜽 =
𝟏
𝐜𝐨𝐬 𝜽
=
−𝟓
𝟑
.
Hence, 𝐜𝐨𝐬 𝜽 =
−𝟑
𝟓
; 𝐭𝐚𝐧 𝜽 =
𝟒
𝟑
; 𝐜𝐨𝐭 𝜽 =
𝟑
𝟒
; 𝐬𝐞𝐜 𝜽 =
−𝟓
𝟑
and 𝐜𝐨𝐬𝐞𝐜 𝜽 =
−𝟓
𝟒
.

Find the value of
(i) 𝐬𝐢𝐧 (
𝟐𝟓𝝅
𝟑
) = 𝐬𝐢𝐧 (𝟖𝝅 +
𝝅
𝟑
) = 𝐬𝒊𝐧
𝝅
𝟑
[ 𝐬𝒊𝐧(𝟐𝒏𝝅 + 𝜽) = 𝐬𝐢𝐧 𝜽]=
√𝟑
𝟐
.
(ii) 𝐜𝐨𝐬 (
𝟒𝟏𝝅
𝟒
) = 𝐜𝐨𝐬 (𝟏𝟎𝝅 +
𝝅
𝟒
) = 𝐜𝐨𝐬
𝝅
𝟒
[ 𝐜𝐨𝐬 (𝟐𝒏𝝅 + 𝜽) = 𝐜𝐨𝐬 𝜽] =
𝟏
√𝟐
.
(iii) 𝐭𝐚𝐧 (−
𝟏𝟔𝝅
𝟑
) = − 𝐭𝐚𝐧
𝟏𝟔𝝅
𝟑
[ 𝐭𝐚𝐧 (−𝜽) = − 𝐭𝐚𝐧 𝜽] = − 𝐭𝐚𝐧 (𝟓𝝅 +
𝝅
𝟑
)
= − 𝐭𝐚𝐧
𝝅
𝟑
[ 𝐭𝐚𝐧 (𝒏𝝅 + 𝜽) = 𝐭𝐚𝐧 𝜽] = −√𝟑.
(iv) 𝐜𝐨𝐭 (
𝟐𝟗𝝅
𝟒
) = 𝐜𝐨𝐭 (𝟕𝝅 +
𝝅
𝟒
) = 𝐜𝐨𝐭
𝝅
𝟒
[ 𝐜𝐨𝐭 (𝒏𝝅 + 𝜽) = 𝐜𝐨𝐭 𝜽] = 𝟏.

(v) 𝐬𝐞𝐜 (−
𝟏𝟗𝝅
𝟑
) = 𝐬𝐞𝐜 (
𝟏𝟗𝝅
𝟑
) [.⋅ 𝐬𝐞𝐜 (−𝜽) = 𝐬𝐞𝐜 𝜽] = 𝐬𝐞𝐜 (𝟔𝝅 +
𝝅
𝟑
) =
𝐬𝐞𝐜 (
𝝅
𝟑
) [ 𝐬𝐞𝐜 (𝟐𝒏𝝅 + 𝜽) = 𝐬𝐞𝐜 𝜽] = 𝟐.
(vi) 𝐜𝐨𝐬𝐞𝐜 (
−𝟑𝟑𝝅
𝟒
) = − 𝐜𝐨𝐬𝐞𝐜
𝟑𝟑𝝅
𝟒
[.⋅ 𝐜𝐨𝐬𝐞𝐜 (−𝜽) = − 𝐜𝐨𝐬𝐞𝐜 𝜽]
= − 𝐜𝐨𝐬𝐞𝐜 (𝟖𝝅 +
𝝅
𝟒
𝝅
𝟒
[ 𝐜𝐨𝐬𝐞𝐜 (𝟐𝒏𝝅 + 𝜽) = 𝐜𝐨𝐬𝐞𝐜 𝜽] = −√𝟐.

Find the value of (i) 𝐬𝐢𝐧 (𝟕𝟔𝟓∘) (ii) 𝐜𝐨𝐬𝐞𝐜 (−𝟏𝟏𝟏𝟎∘) (iii) 𝐜𝐨𝐭 (−𝟔𝟎𝟎∘)
(i) 𝟏𝟖𝟎∘
= 𝝅𝐜
⇒ 𝟕𝟔𝟓∘
= (
𝝅
𝟏𝟖𝟎
× 𝟕𝟔𝟓)
𝐜
= (
𝟏𝟕𝝅
𝟒
)
𝐜
𝐬𝐢𝐧 (𝟕𝟔𝟓∘) = 𝐬𝒊𝒏 (
𝟏𝟕𝝅
𝟒
) = 𝐬𝐢𝐧 (𝟒𝝅 +
𝝅
𝟒
) = 𝐬𝐢𝐧
𝝅
𝟒
[ 𝐬𝐢𝐧 (𝟐𝒏𝝅 + 𝜽) = 𝐬𝐢𝐧 𝛉] =
𝟏
√𝟐
.
(ii) 𝟏𝟖𝟎∘
= 𝝅𝐜
⇒ (𝟏𝟏𝟏𝟎) ∘
= (
𝝅
𝟏𝟖𝟎
× 𝟏𝟏𝟏𝟎)
𝐜
= (
𝟑𝟕𝝅
𝟔
)
𝐜
𝐜𝐨𝐬𝐞𝐜 (−𝟏𝟏𝟏𝟎∘) = − 𝐜𝐨𝐬𝐞𝐜 (𝟏𝟏𝟏𝟎∘) [ 𝐜𝐨𝐬𝐞𝐜 (−𝜽) = − 𝐜𝐨𝐬𝐞𝐜 𝜽] = − 𝐜𝐨𝐬𝐞𝐜 (
𝟑𝟕𝝅
𝟔
)
= − 𝐜𝐨𝐬𝐞𝐜 (𝟔𝝅 +
𝝅
𝟔
𝝅
𝟔
[ 𝐜𝐨𝐬𝐞𝐜 (𝟐𝒏𝝅 + 𝜽) = 𝐜𝐨𝐬𝐞𝐜 𝜽] = −𝟐.

(iii) 𝟏𝟖𝟎∘
= 𝝅𝐜
⇒ 𝟔𝟎𝟎∘
= (
𝝅
𝟏𝟖𝟎
× 𝟔𝟎𝟎)
𝐜
= (
𝟏𝟎𝝅
𝟑
)
𝐜
𝐜𝐨𝐭 (−𝟔𝟎𝟎∘) = − 𝐜𝐨𝐭 𝟔𝟎𝟎∘ [ 𝐜𝐨𝐭 (−𝜽) = − 𝐜𝐨𝐭 𝜽] = − 𝐜𝐨𝐭 (
𝟏𝟎𝝅
𝟑
)
= − 𝐜𝐨𝐭 (𝟑𝝅 +
𝝅
𝟑
) = − 𝐜𝐨𝐭
𝝅
𝟑
= −
𝟏
√𝟑
. [ 𝐜𝐨𝐭 (𝒏𝝅 + 𝜽) = 𝐜𝐨𝐭 𝜽]

Find the value of
(i) 𝐜𝐨𝐬 𝟏𝟓𝝅 = 𝐜𝐨𝐬 (𝟏𝟒𝝅 + 𝝅) = 𝐜𝐨𝐬 𝝅 [ 𝐜𝐨𝐬 (𝟐𝒏𝝅 + 𝜽) = 𝐜𝐨𝐬 𝜽] = −𝟏.
(ii) 𝐬𝐢𝐧 𝟏𝟔𝝅 = 𝐬𝐦
̇ (𝟏𝟔𝝅 + 𝟎) = 𝐬𝐢𝐧 𝟎∘ [ 𝐬𝐢𝐧 (𝟐𝒏𝝅 + 𝜽) = 𝐬𝐢𝐧 𝜽] = 𝟎.
(iii) 𝐜𝐨𝐬 (−𝝅) = 𝐜𝐨𝐬 𝝅 [ 𝐜𝐨𝐬 (−𝜽) = 𝐜𝐨𝐬 𝜽] = −𝟏.
(iv) 𝐬𝐢𝐧 𝟓𝝅 = 𝐬𝐢𝐧 (𝟒𝝅 + 𝝅) = 𝐬𝐢𝐧𝝅 [ 𝐬𝐢𝐧(𝟐𝒏𝝅 + 𝜽) = 𝐬𝐢𝐧 𝜽]= 𝟎.
(v) 𝐭𝐚𝐧
𝟓𝝅
𝟒
= 𝐭𝐚𝐧 (𝝅 +
𝝅
𝟒
) = 𝐭𝐚𝐧
𝝅
𝟒
[ 𝐭𝐚𝐧 (𝒏𝝅 + 𝜽) = 𝐭𝐚𝐧 𝜽] = 𝟏.
(vi) 𝐬𝐞𝐜 𝟔𝝅 = 𝐬𝐞𝐜 (𝟔𝝅 + 𝟎) = 𝐬𝐞𝐜 𝟎∘ [ 𝐬𝐞𝐜 (𝟐𝒏𝝅 + 𝜽) = 𝐬𝐞𝐜 𝜽] = 𝟏.

Prove that 𝐬𝐢𝐧𝟐 𝝅
𝟔
+ 𝐜𝐨𝐬𝟐 𝝅
𝟑
—𝒕𝒂𝒏𝟐 𝝅
𝟒
= −
𝟏
𝟐
.
LHS = 𝐬𝐢𝐧𝟐 𝝅
𝟔
+ 𝐜𝐨𝐬𝟐 𝝅
𝟑
—𝒕𝒂𝒏𝟐 𝝅
𝟒
= (
𝟏
𝟐
)
𝟐
+ (
𝟏
𝟐
)
𝟐
− 𝟏𝟐
[⋅.⋅ 𝐬i𝐧
𝝅
𝟔
=
𝟏
𝟐
, 𝐜𝐨𝐬
𝝅
𝟑
=
𝟏
𝟐
and 𝐭𝐚𝐧
𝝅
𝟒
= 𝟏]
= (
𝟏
𝟒
+
𝟏
𝟒
− 𝟏) = −
𝟏
𝟐
= 𝐑𝐇𝐒.

Class 11_Chapter 3_Lecture_2

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Class 11_Chapter 3_Lecture_2