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Question: Prove that \[\dfrac{{{\cot }^{2}}A\left( \sec A-1 \right)}{1+\sin A}={{\sec }^{2}}A\left( \dfrac{1-\...

Prove that cot⁡2A(sec⁡A−1)1+sin⁡A=sec⁡2A(1−sin⁡A1+sec⁡A)\dfrac{{{\cot }^{2}}A\left( \sec A-1 \right)}{1+\sin A}={{\sec }^{2}}A\left( \dfrac{1-\sin A}{1+\sec A} \right).

Explanation

Solution

Hint: Take LHS individually and by using Trigonometric Identities Trigonometric ratios simplify it. Similarly, take RHS and simplify it.

“Complete step-by-step answer:”
Given the LHS=cot⁡2A(sec⁡A−1)1+sin⁡ALHS=\dfrac{{{\cot }^{2}}A\left( \sec A-1 \right)}{1+\sin A}
We know that cot⁡A=cos⁡Asin⁡A\cot A=\dfrac{\cos A}{\sin A}and sec⁡A=1cos⁡A\sec A=\dfrac{1}{\cos A}
∴LHS=cos⁡2Asin⁡2A[1cos⁡A−1]1+sin⁡A=cos⁡2Asin⁡2A[1−cos⁡Acos⁡A]1+sin⁡A\therefore LHS=\dfrac{\dfrac{{{\cos }^{2}}A}{{{\sin }^{2}}A}\left[ \dfrac{1}{\cos A}-1 \right]}{1+\sin A}=\dfrac{\dfrac{{{\cos }^{2}}A}{{{\sin }^{2}}A}\left[ \dfrac{1-\cos A}{\cos A} \right]}{1+\sin A}
Cancel out cos⁡A\cos Afrom numerator and denominator.
LHS=cos⁡Asin⁡2A(1−cos⁡A)1+sin⁡ALHS=\dfrac{\dfrac{\cos A}{{{\sin }^{2}}A}\left( 1-\cos A \right)}{1+\sin A}
We know, sin⁡2A+cos⁡2A=1{{\sin }^{2}}A+{{\cos }^{2}}A=1

& \Rightarrow {{\sin }^{2}}A=1-{{\cos }^{2}}A \\\ & 1-{{\cos }^{2}}A=\left( 1-\cos A \right)\left( 1+\cos A \right) \\\ & \therefore LHS=\dfrac{\dfrac{\cos A}{\left( 1-{{\cos }^{2}}A \right)}\left( 1-\cos A \right)}{1+\sin A} \\\ & =\dfrac{\cos A\left( 1-\cos A \right)}{\left( 1-\cos A \right)\left( 1+\cos A \right)\left( 1+\sin A \right)} \\\ \end{aligned}$$ Cancel out $$\left( 1-\cos A \right)$$from numerator and denominator. $$=\dfrac{\cos A}{\left( 1+\cos A \right)\left( 1+\sin A \right)}-\left( 1 \right)$$ Now take $$RHS={{\sec }^{2}}A\left( \dfrac{1-\sin A}{1+\sec A} \right)$$ We know, $$\sec A=\dfrac{1}{\cos A}$$ $$\begin{aligned} & \Rightarrow RHS=\dfrac{1}{{{\cos }^{2}}A}\left[ \dfrac{1-\sin A}{1+\dfrac{1}{\cos A}} \right] \\\ & =\dfrac{1}{{{\cos }^{2}}A}\left[ \dfrac{1-\sin A}{\dfrac{\cos A+1}{\cos A}} \right] \\\ & =\dfrac{1}{{{\cos }^{2}}A}\left[ \dfrac{\cos A\left( 1-\sin A \right)}{1+\cos A} \right] \\\ \end{aligned}$$ We know, $${{\sin }^{2}}A+{{\cos }^{2}}A=1$$ $$\begin{aligned} & \Rightarrow {{\cos }^{2}}A=1-{{\sin }^{2}}A \\\ & \left( 1-{{\sin }^{2}}A \right)=\left( 1-\sin A \right)\left( 1+\sin A \right) \\\ & \therefore RHS=\dfrac{1}{\left( 1-\sin A \right)\left( 1+\sin A \right)}\left[ \dfrac{\cos A\left( 1-\sin A \right)}{1+\cos A} \right] \\\ \end{aligned}$$ Cancel out $$\left( 1-\sin A \right)$$ from numerator and denominator. $$RHS=\dfrac{\cos A}{\left( 1+\cos A \right)\left( 1+\sin A \right)}-(2)$$ Now (1) = (2) Which shows that LHS = RHS. Note: By solving LHS and RHS, we simplify both to $$\dfrac{\cos A}{\left( 1+\cos A \right)\left( 1+\sin A \right)}$$which shows that LHS = RHS.