Trigonometric Ratios Of Complementary Angles
Consider the following figure.
Here, a right-angled triangle ABC has been shown. In this triangle, suppose that the value of
sin C is .
Can we find the value of cos A?
We can use these relations for simplifying the given expression.
For example: Let us express sec 55°− cosec 89° in terms of trigonometric ratios of angles between 0° and 45°.
Now, how can we do so? Let us see.
Since 55° and 35° are complementary angles and also 89° and 1° are complementary angles, we can write 55° as (90° − 35°) and 89° as (90° − 1°).
Therefore,
sec 55° − cosec 89° = sec (90° − 35°) − cosec (90° − 1°)
= cosec 35° − sec 1°
[ sec (90° – A) = cosec A and cosec (90° – A) = sec A]
∴ sec 55° − cosec 89°= cosec 35° − sec 1°
Let us now solve some more examples involving trigonometric ratios of complementary angles.
Example 1: Find the value of sin 53° – cos 37°.
Solution:
We know that 53° and 37° are complementary angles as
53° + 37° = 90°
∴ We can write 37° as (90° – 53°).
∴ sin 53° – cos 37° = sin 53° – cos (90° – 53°)
= sin 53° – sin 53°
[ cos (90° – A) = sin A]
= 0
Thus, the value of (sin 53° – cos 37°) is 0.
Example 2: Evaluate
Solution:
Here, 27° and 63° are complementary angles as 27° + 63° = 90°
∴ We can write 27° = 90° – 63°
Now,
Example 3: Prove that tan 2A = cot 3A, when A = 18°.
Solution:
When A = 18°,
L.H.S = tan 2A = tan (2 × 18°)
= tan 36°
R.H.S = cot 3A = cot (3 × 18)
= cot 54°
54° and 36° are complementary angles.
∴ We can write 54° as 90° – 36°.
Therefore, cot 3A = cot 54°
= cot (90° – 36°)
= tan 36° [ cot (90° – A) = tan A]
∴ L.H.S = R.H.S = tan 36°
∴ tan 2A = cot 3A
Example 4: If sin A = cos A, then prove that A = 45°.
Solution:
It is given that sin A = cos A
⇒ sin A = sin (90° – A)
[ sin (90° – A) = cos A]
⇒ A = 90° – A
⇒ 2A = 90°
Hence, proved
Example 5: If P, Q, and R are interior angles of a triangle PQR, which is right-angled at Q, then show that
Solution:
Now, P, Q, and R are the interior angles of the triangle PQR. Therefore, their sum should be 180°.
∴ P + R = 180 – Q
Now, consider the L.H.S. =
=
=
= [ cot (90° – A) = tan A]
= R.H.S.
∴
Hence, proved
Example 6: Prove that
tan1°× tan 2°× tan 3° … tan 87°× tan 88°× tan 89° = 1
Solution:
Here, 1° and 89° are complementary angles as 1° + 89° = 90°
Therefore, we can write 89° = 90° – 1°
Similarly, 88° = 90° – 2°
87° = 90° – 3°
46° = 90° – 44° and so on
Now, the L.H.S is
tan 1° × tan 2° × tan 3° … tan 44° × tan 45° × tan 46° … tan 87° × tan 88° × tan 89°
= tan 1° × tan 2° × tan 3° … tan 44° × tan 45° × tan (90° – 44°) … tan (90° – 3°) tan (90° – 2°)
tan (90° – 1°)
= tan 1° × tan 2° × tan 3° … tan 44° × tan 45° × cot 44° … cot 3° × cot 2° × cot 1°
[ tan (90° – A) = cot A]
= tan 45°
= 1
= R.H.S
∴ tan1° × tan 2° × tan 3° … tan 87° × tan 88° × tan 89° = 1
Hence, proved