Question

11. Prove that: (cosec A-sin A) (sec A- cos A) (tan A + cot A)=1​

Answer 1

Answer:

Proved below!

Step-by-step explanation:

To prove:

(cosec A-sin A) (sec A- cos A) (tan A + cot A)=1

L.H.S = (cosec A-sin A) (sec A- cos A) (tan A + cot A)

=> (tanA + cotA)(cosecA - sinA)(secA, - cosA)

=> $(tanA + \frac{1}{tanA})(\frac{1}{sinA} - sinA)(\frac{1}{cosA} - cosA)$

=> $(\frac{tan^2A + 1}{tanA})(\frac{1 - sin^2A}{sinA})(\frac{1 - cos^2A}{cosA})$

=> We know that,

tan²A + 1 = sec²A

1 - sin²A = cos²A

1 - cos²A = sin²A

=> $(\frac{sec^2A}{tanA})(\frac{cos^2A}{sin^2A})(\frac{sin^2A}{cosA})$

[∵ secA = 1/cosA]

=> $\therefore \frac{\cancel{sec^2A}}{tanA} \times \frac{\cancel{cos^2A}}{\cancel{sinA}} \times \frac{sin^{\cancel{2}}A}{cosA}$

=> $\frac{\cancel{sinA}}{\cancel{cosA}} \times \frac{\cancel{1}}{\cancel{tanA}}$

=> 1 = R.H.S

=> L.H.S = R.H.S

=> Hence, the proof.