Question

Please solve.this question​

Answer 1

Answer:

See explanation

Step-by-step explanation:

Prove equality

$(1-\cot A)^2+(\tan A-1)^2=4\csc 2A(\csc 2A-1)$

Consider left and right parts separately.

Left part:

$(1-\cot A)^2+(\tan A-1)^2\\ \\=1-2\cot A+\cot^2 A+\tan^2 A-2\tan A+1\\ \\=(1+\cot^2 A)+(1+\tan^2 A)-2(\cot A+\tan A)\\ \\=\dfrac{1}{\sin^2 A}+\dfrac{1}{\cos ^2A}-2\dfrac{\cos^2 A+\sin^2 A}{\cos A\sin A}\\ \\=\dfrac{1}{\sin^2 A}+\dfrac{1}{\cos ^2A}-\dfrac{2}{\cos A\sin A}\\ \\=\dfrac{\cos^2A+\sin^2A-2\cos A\sin A}{\cos ^2A\sin ^2A}\\ \\=\dfrac{1-\sin 2A}{\frac{1}{4}\cdot 4\cdot \cos^2A\sin^2 A}\\ \\=\dfrac{1-\sin 2A}{\frac{1}{4}\cdot \sin^22 A}\\ \\=4\dfrac{1-\sin 2A}{\sin^22 A}$

Right part:

$\csc 2A=\dfrac{1}{\sin 2A}$

Hence

$4\csc 2A(\csc 2A-1)\\ \\=4\cdot \dfrac{1}{\sin 2A}\cdot \left(\dfrac{1}{\sin 2A}-1\right)\\ \\=4\cdot \dfrac{1}{\sin 2A}\cdot \dfrac{1-\sin 2A}{\sin 2A}\\ \\=4\dfrac{1-\sin 2A}{\sin^2 2A}$

Since left and right parts are the same, the equality is true.