It is a trigonometric expression which contains sine and cosine functions with complementary angles. This trigonometric problem can be simplified in two methods by using complementary angle trigonometric identities.

In this method, all trigonometric functions which contain complementary angles are simplified firstly by using cofunction identities.

$\dfrac{\cos{(90^°-\theta)}}{1+\sin{(90^°-\theta)}}$ $+$ $\dfrac{1+\sin{(90^°-\theta)}}{\cos{(90^°-\theta)}}$

The sin of angle is equal to cos of complementary angle as per cofunction identity of sin function and the cos of angle is equal to sin of complementary angle as per cofunction identity of cos function.

$= \,\,\,$ $\dfrac{\sin{\theta}}{1+\cos{\theta}}$ $+$ $\dfrac{1+\cos{\theta}}{\sin{\theta}}$

The two terms can be simplified by using least common multiple method for making them as a single term trigonometric expression.

$= \,\,\,$ $\dfrac{\sin{\theta} \times \sin{\theta} + {(1+\cos{\theta})}{(1+\cos{\theta})}}{\sin{\theta}{(1+\cos{\theta})}}$

$= \,\,\,$ $\dfrac{\sin^2{\theta}+{(1+\cos{\theta})}^2}{\sin{\theta}{(1+\cos{\theta})}}$

The second term in the numerator is a square of sum of two terms and it can be expanded by using square of sum of two terms formula like ${(a+b)}^2$.

$= \,\,\,$ $\dfrac{\sin^2{\theta}+1^2+\cos^2{\theta}+2 \times 1 \times \cos{\theta}}{\sin{\theta}{(1+\cos{\theta})}}$

$= \,\,\,$ $\dfrac{\sin^2{\theta}+1+\cos^2{\theta}+2\cos{\theta}}{\sin{\theta}{(1+\cos{\theta})}}$

$= \,\,\,$ $\dfrac{1+\sin^2{\theta}+\cos^2{\theta}+2\cos{\theta}}{\sin{\theta}{(1+\cos{\theta})}}$

The sum of squares of sin and cos functions is one at an angle as per Pythagorean identity of sin and cos functions.

$= \,\,\,$ $\dfrac{1+1+2\cos{\theta}}{\sin{\theta}{(1+\cos{\theta})}}$

$= \,\,\,$ $\dfrac{2+2\cos{\theta}}{\sin{\theta}{(1+\cos{\theta})}}$

$= \,\,\,$ $\dfrac{2(1+\cos{\theta})}{\sin{\theta}{(1+\cos{\theta})}}$

$= \,\,\,$ $\require{\cancel} \dfrac{2\cancel{(1+\cos{\theta})}}{\sin{\theta}\cancel{(1+\cos{\theta})}}$

$= \,\,\,$ $\dfrac{2}{\sin{\theta}}$

According to reciprocal identity of sin function, the reciprocal of sin function is equal to cosecant function.

$= \,\,\,$ $2\csc{\theta}$

It is a method for those who have basic knowledge on trigonometry and have a confidence about dealing trigonometric functions which contain complementary angles. In this method, no trigonometric ratio with complementary angle is transformed till last step.

$\dfrac{\cos{(90^°-\theta)}}{1+\sin{(90^°-\theta)}}$ $+$ $\dfrac{1+\sin{(90^°-\theta)}}{\cos{(90^°-\theta)}}$

Firstly, combine the terms in the form of trigonometric function by least common multiple method.

$= \,\,\,$ $\dfrac{\cos{(90^°-\theta)} \times \cos{(90^°-\theta)} + {(1+\sin{(90^°-\theta)})}{(1+\sin{(90^°-\theta)})}}{\cos{(90^°-\theta)}{(1+\sin{(90^°-\theta)})}}$

$= \,\,\,$ $\dfrac{\cos^2{(90^°-\theta)} + {(1+\sin{(90^°-\theta)})}^2}{\cos{(90^°-\theta)}{(1+\sin{(90^°-\theta)})}}$

Now, expand the square of sum of two terms in the numerator.

$= \,\,\,$ $\dfrac{\cos^2{(90^°-\theta)} + 1^2+\sin^2{(90^°-\theta)}+2 \times 1 \times \sin{(90^°-\theta)}}{\cos{(90^°-\theta)}{(1+\sin{(90^°-\theta)})}}$

$= \,\,\,$ $\dfrac{\cos^2{(90^°-\theta)} + 1 +\sin^2{(90^°-\theta)}+2\sin{(90^°-\theta)}}{\cos{(90^°-\theta)}{(1+\sin{(90^°-\theta)})}}$

$= \,\,\,$ $\dfrac{1+\cos^2{(90^°-\theta)}+\sin^2{(90^°-\theta)}+2\sin{(90^°-\theta)}}{\cos{(90^°-\theta)}{(1+\sin{(90^°-\theta)})}}$

As per Pythagorean identity of sin and cos functions, the sum of squares of sin and cos functions at an angle is equal to one.

$= \,\,\,$ $\dfrac{1+1+2\sin{(90^°-\theta)}}{\cos{(90^°-\theta)}{(1+\sin{(90^°-\theta)})}}$

$= \,\,\,$ $\dfrac{2+2\sin{(90^°-\theta)}}{\cos{(90^°-\theta)}{(1+\sin{(90^°-\theta)})}}$

$= \,\,\,$ $\dfrac{2(1+\sin{(90^°-\theta)})}{\cos{(90^°-\theta)}{(1+\sin{(90^°-\theta)})}}$

$= \,\,\,$ $\require{cancel} \dfrac{2\cancel{(1+\sin{(90^°-\theta)})}}{\cos{(90^°-\theta)}\cancel{(1+\sin{(90^°-\theta)})}}$

$= \,\,\,$ $\dfrac{2}{\cos{(90^°-\theta)}}$

The reciprocal of cos of angle is equal to secant of angle as per reciprocal rule of cos function.

$= \,\,\,$ $2\sec{(90^°-\theta)}$

Finally, the secant of angle is equal to cosecant of complementary angle as per complementary angle identity of secant function.

$= \,\,\,$ $2\csc{\theta}$

Latest Math Topics

Aug 31, 2024

Aug 07, 2024

Jul 24, 2024

Dec 13, 2023

Latest Math Problems

Sep 04, 2024

Jan 30, 2024

Oct 15, 2023

Copyright © 2012 - 2023 Math Doubts, All Rights Reserved