Question

Please solve.this question​

Answer 1

See explanation

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)\\ \\=(1)/(\sin^2 A)+(1)/(\cos ^2A)-2(\cos^2 A+\sin^2 A)/(\cos A\sin A)\\ \\=(1)/(\sin^2 A)+(1)/(\cos ^2A)-(2)/(\cos A\sin A)\\ \\=(\cos^2A+\sin^2A-2\cos A\sin A)/(\cos ^2A\sin ^2A)\\ \\=(1-\sin 2A)/((1)/(4)\cdot 4\cdot \cos^2A\sin^2 A)\\ \\=(1-\sin 2A)/((1)/(4)\cdot \sin^22 A)\\ \\=4(1-\sin 2A)/(\sin^22 A)`

Right part:

`\csc 2A=(1)/(\sin 2A)`

Hence

`4\csc 2A(\csc 2A-1)\\ \\=4\cdot (1)/(\sin 2A)\cdot \left((1)/(\sin 2A)-1\right)\\ \\=4\cdot (1)/(\sin 2A)\cdot (1-\sin 2A)/(\sin 2A)\\ \\=4(1-\sin 2A)/(\sin^2 2A)`

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