Trigonometric Ratios Of Complementary Angles

Consider the following figure.

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/180/456/952/10.1.8.3.1_ok_SS_html_m51d5fe01.png

Here, a right-angled triangle ABC has been shown. In this triangle, suppose that the value of

sin C is https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/180/456/952/10.1.8.3.1_ok_SS_html_m109a95fe.gif.

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°

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/180/456/952/10.1.8.3.1_ok_SS_html_6290838.gif[ 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°

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/180/456/952/10.1.8.3.1_ok_SS_html_6290838.gif[ cos (90° – A) = sin A]

= 0

Thus, the value of (sin 53° – cos 37°) is 0.

Example 2: Evaluatehttps://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/180/456/952/10.1.8.3.1_ok_SS_html_m51175726.gif

Solution:

Here, 27° and 63° are complementary angles as 27° + 63° = 90°

∴ We can write 27° = 90° – 63°

Now,

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/180/456/952/10.1.8.3.1_ok_SS_html_m24b6cbb6.gif

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° [ https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/180/456/952/10.1.8.3.1_ok_SS_html_6290838.gifcot (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)

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/180/456/952/10.1.8.3.1_ok_SS_html_6290838.gif[ sin (90° – A) = cos A]

⇒ A = 90° – A

⇒ 2A = 90°

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/180/456/952/10.1.8.3.1_ok_SS_html_5c687438.gif

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

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/180/456/952/10.1.8.3.1_ok_SS_html_m3842ccd9.gif

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. =https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/180/456/952/10.1.8.3.1_ok_SS_html_m568c63f.gif

=https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/180/456/952/10.1.8.3.1_ok_SS_html_106f6d06.gif

=https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/180/456/952/10.1.8.3.1_ok_SS_html_2eae9090.gif

=https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/180/456/952/10.1.8.3.1_ok_SS_html_21cfc021.gif [ https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/180/456/952/10.1.8.3.1_ok_SS_html_6290838.gifcot (90° – A) = tan A]

= R.H.S.

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/180/456/952/10.1.8.3.1_ok_SS_html_m6c36f94d.gif

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°

[ https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/180/456/952/10.1.8.3.1_ok_SS_html_6290838.giftan (90° – A) = cot A]

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/180/456/952/10.1.8.3.1_ok_SS_html_m70db81d8.gif

= tan 45°

= 1

= R.H.S

∴ tan1° × tan 2° × tan 3° … tan 87° × tan 88° × tan 89° = 1

Hence, proved