Solveeit Logo
Login

Question

Question: Solve\[\dfrac{{\left( {cosA - sinA + 1} \right)}}{{\left( {cosA + sinA - 1} \right)}} = \csc A + cot...

Solve(cosA−sinA+1)(cosA+sinA−1)=csc⁡A+cotA\dfrac{{\left( {cosA - sinA + 1} \right)}}{{\left( {cosA + sinA - 1} \right)}} = \csc A + cotA, using the identity csc⁡2A=1+cot2A{\csc ^2}A = 1 + co{t^2}A

Explanation

Solution

To solve these questions, we use the relationship between trigonometry functions to simplify the left-hand side of the equation, and then we use the given identity to equate the given left-hand side to the right-hand side.

Complete step-by-step answer:
Firstly we will take the left-hand side of the question
L.H.S.:(cosA−sinA+1)(cosA+sinA−1)\dfrac{{\left( {cosA - sinA + 1} \right)}}{{\left( {cosA + sinA - 1} \right)}}
Take sin⁡A\sin Acommon from numerator and denominator
=sin⁡A(cos⁡Asin⁡A−sin⁡Asin⁡A+1sin⁡A)sin⁡A(cos⁡Asin⁡A+sin⁡Asin⁡A−1sin⁡A)......(1)= \dfrac{{\sin A(\dfrac{{\cos A}}{{\sin A}} - \dfrac{{\sin A}}{{\sin A}} + \dfrac{1}{{\sin A}})}}{{\sin A(\dfrac{{\cos A}}{{\sin A}} + \dfrac{{\sin A}}{{\sin A}} - \dfrac{1}{{\sin A}})}}......(1)
Cancel out sin⁡A\sin Afrom numerator and denominator, as we know that cos⁡Asin⁡A=cot⁡A\dfrac{{\cos A}}{{\sin A}} = \cot Aand 1sin⁡A=csc⁡A\dfrac{1}{{\sin A}} = \csc Awe use these identities in above equation (1)(1)
=cot⁡A−1+csc⁡Acot⁡A+1−csc⁡A......(2)= \dfrac{{\cot A - 1 + \csc A}}{{\cot A + 1 - \csc A}}......(2)
Now, csc⁡2A=1+cot2A{\csc ^2}A = 1 + co{t^2}A identity is given to use
Which we simplify as, csc⁡2A−cot⁡2A=1{\csc ^2}A - {\cot ^2}A = 1
We will put value of 11 in equation (2)(2)
=cot⁡A−(csc⁡2A−cot⁡2A)+csc⁡Acot⁡A+1−csc⁡A= \dfrac{{\cot A - ({{\csc }^2}A - {{\cot }^2}A) + \csc A}}{{\cot A + 1 - \csc A}}
Here, we will use a2−b2=(a−b)(a+b){a^2} - {b^2} = (a - b)(a + b)formula in above equation
We put a=csc⁡Aa = \csc Aand b=cot⁡Ab = \cot Ain above formula
=cot⁡A+csc⁡A−(csc⁡A−cot⁡A)(csc⁡A+cot⁡A)cot⁡A+1−csc⁡A= \dfrac{{\cot A + \csc A - (\csc A - \cot A)(\csc A + \cot A)}}{{\cot A + 1 - \csc A}}
=(cot⁡A+csc⁡A)(1−csc⁡A+cot⁡A)cot⁡A+1−csc⁡A= \dfrac{{(\cot A + \csc A)(1 - \csc A + \cot A)}}{{\cot A + 1 - \csc A}}(As we seen in equation that (1−csc⁡A+cot⁡A)(1 - \csc A + \cot A)is present in both numerator and denominator, we cancel them)
=cot⁡A+csc⁡A= \cot A + \csc A=R.H.S.
Hence, we proved that L.H.S. = R.H.S.

Additional information:
Trigonometry functions: In trigonometry, these functions tell the relation between the angles and the ratio of two sides of the right-angle triangle. It is also known as the circular function or angle function.
Trigonometry identities: Trigonometric identities are balances that involve trigonometric functions. They are true for every value of the occurring variables where both sides of the equation are defined. Geometrically, these are identities involving certain functions of one or more angles.

Note: We should remember the relationships between sin⁡A,cos⁡A,cot⁡A\sin A,\cos A,\cot Aand csc⁡A\csc A trigonometry functions as these terms used in the question and a2−b2=(a−b)(a+b){a^2} - {b^2} = (a - b)(a + b) identity, to simplify complex equations.