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Question: How do you solve \(\dfrac{{1 + \sin x + \cos x}}{{1 + \sin x - \cos x}} = \dfrac{{1 + \cos x}}{{\sin...

How do you solve 1+sin⁡x+cos⁡x1+sin⁡x−cos⁡x=1+cos⁡xsin⁡x\dfrac{{1 + \sin x + \cos x}}{{1 + \sin x - \cos x}} = \dfrac{{1 + \cos x}}{{\sin x}}?

Explanation

Solution

Hint : Here, in the given question, we need to prove that 1+sin⁡x+cos⁡x1+sin⁡x−cos⁡x=1+cos⁡xsin⁡x\dfrac{{1 + \sin x + \cos x}}{{1 + \sin x - \cos x}} = \dfrac{{1 + \cos x}}{{\sin x}}. We will first multiply and divide LHS by sin⁡x\sin x. As we can see that in the right-hand side, the numerator value is in the terms of 1+cos⁡x1 + \cos x, so we will try to convert sin⁡e\sin e function into cos⁡ine\cos ine using identities. After that we will take out the common terms and cancel out the common terms to get our required answer.
Formulae used :
I.sin⁡2x=1−cos⁡2x{\sin ^2}x = 1 - {\cos ^2}x
II.tan⁡2x=sec⁡2x−1{\tan ^2}x = {\sec ^2}x - 1
III.cos⁡ec2x=1+cot⁡2x\cos e{c^2}x = 1 + {\cot ^2}x

Complete step-by-step answer :
We have to prove that, 1+sin⁡x+cos⁡x1+sin⁡x−cos⁡x=1+cos⁡xsin⁡x\dfrac{{1 + \sin x + \cos x}}{{1 + \sin x - \cos x}} = \dfrac{{1 + \cos x}}{{\sin x}}
L.H.S, 1+sin⁡x+cos⁡x1+sin⁡x−cos⁡x\dfrac{{1 + \sin x + \cos x}}{{1 + \sin x - \cos x}}
Multiply and divide by sin⁡x\sin x
⇒sin⁡xsin⁡x×1+sin⁡x+cos⁡x1+sin⁡x−cos⁡x\Rightarrow \dfrac{{\sin x}}{{\sin x}} \times \dfrac{{1 + \sin x + \cos x}}{{1 + \sin x - \cos x}}
On multiplication of sin⁡x\sin x with numerator, we get
⇒1sin⁡x×[sin⁡x+sin⁡x.sin⁡x+sin⁡x.cos⁡x1+sin⁡x−cos⁡x]\Rightarrow \dfrac{1}{{\sin x}} \times \left[ {\dfrac{{\sin x + \sin x.\sin x + \sin x.\cos x}}{{1 + \sin x - \cos x}}} \right]
⇒1sin⁡x×[sin⁡x+sin⁡x2+sin⁡x.cos⁡x1+sin⁡x−cos⁡x]\Rightarrow \dfrac{1}{{\sin x}} \times \left[ {\dfrac{{\sin x + \sin {x^2} + \sin x.\cos x}}{{1 + \sin x - \cos x}}} \right]
As we know, sin⁡2x=1−cos⁡2x{\sin ^2}x = 1 - {\cos ^2}x. Therefore, we get
⇒1sin⁡x×[sin⁡x+sin⁡x.cos⁡x+12−cos⁡2x1+sin⁡x−cos⁡x]\Rightarrow \dfrac{1}{{\sin x}} \times \left[ {\dfrac{{\sin x + \sin x.\cos x + {1^2} - {{\cos }^2}x}}{{1 + \sin x - \cos x}}} \right]
On taking sin⁡x\sin x as a common factor and expanding 12−cos⁡2x{1^2} - {\cos ^2}x, using a2−b2=(a+b)(a−b){a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right), we get
⇒1sin⁡x×[sin⁡x(1+cos⁡x)+(1+cos⁡x)(1−cos⁡x)1+sin⁡x−cos⁡x]\Rightarrow \dfrac{1}{{\sin x}} \times \left[ {\dfrac{{\sin x\left( {1 + \cos x} \right) + \left( {1 + \cos x} \right)\left( {1 - \cos x} \right)}}{{1 + \sin x - \cos x}}} \right]
Now, we will take (1+cos⁡x)\left( {1 + \cos x} \right) as a common term
⇒1sin⁡x×[(1+cos⁡x)(sin⁡x+1−cos⁡x)1+sin⁡x−cos⁡x]\Rightarrow \dfrac{1}{{\sin x}} \times \left[ {\dfrac{{\left( {1 + \cos x} \right)\left( {\sin x + 1 - \cos x} \right)}}{{1 + \sin x - \cos x}}} \right]
On canceling out the common terms, we get
⇒1sin⁡x×[(1+cos⁡x)1]\Rightarrow \dfrac{1}{{\sin x}} \times \left[ {\dfrac{{\left( {1 + \cos x} \right)}}{1}} \right]
This can also be written as,
⇒1+cos⁡xsin⁡x\Rightarrow \dfrac{{1 + \cos x}}{{\sin x}} = RHS
Hence proved, LHS = RHS.

Note : In mathematics, the sine is a trigonometric function of an angle. The sine of an acute angle is defined in the context of a right triangle: for the specified angle, it is the ratio of the length of the side that is opposite that angle, to the length of the longest side of the triangle. Whereas cos is the ratio of the adjacent side to the hypotenuse of a right triangle.