Prove that
(1-tan A)2 + (1 - cot A)2 = (sec A- cosec A).
Answers
Step-by-step explanation:
(1-tanA)²+(1-cotA)²
=(cosA-sinA/cosA)²+(sinA-CosA/sinA)²
=(cosA-sinA/CosA)²+ {-(cosA-sinA/sinA)}²
=(cosA-sinA/cosA)²+(cosA-sinA/sinA)²
=(cosA-sinA)²sin²A+(cosA-sinA)²cos²A/sin²A cos²A
=(cosA-sinA)²(sin²A+cos²A)/sin²A cos²A
=(cosA-sinA)²/sin²A cos²A
=(1/sinA-1/cosA)²
=(cosecA-secA)²
=(secA-cosecA)²
Step-by-step explanation:
How do you prove that sinA(1+tanA)+cosA(1+cotA)=secA+cosecA?
To prove sin A(1+ tan A)+ cos A(1 + cot A) = sec A + cosec A.
LHS = sin A(1+ tan A)+ cos A(1 + cot A)
= sin A + sin^2 A/ cos A + cos A + cos^2 A/ sin A
= sin A + cos A + [sin^3 A + cos^3 A]/sin A cos A
=[ sin^2 A cos A + cos^2 A sin A + sin^3 A + cos^3 A]/sin A cos A
= [ sin^2 A cos A +cos^3 A + cos^2 A sin A + sin^3 A]/sin A cos A
= [cos A (sin^2 A + cos^2 A) + sin A (sin^2 A + cos^2 A)]/sin A cos A
= [cos A +sin A]/sin A cos A
= (1/sin A) + (1/cos A)
= cosec A + sec A = RHS.
Proved.