Question

trigonometry equations​

Answer 1

Step-by-step explanation:

I don't think it's right .. the LHS is not equal to the RHS..

Answer 2

Answer:

LHS 1/(sec A + tan A)

= 1/(1/cos A + sin A/cos A)

= 1/(1 + sin A)/cos A

= cos A/(1+sin A)

multiply both numerator and denominator by (1-sinA)

[cos A *(1-sin A)]/[ (1+sin A)(1-sin A)]

= [ cos A* (1-sinA)]/ ( 1 - sin^2 A)

[ (a-b)(a+b)=a^2-b^2]

= [cos A* (1-sin A)]/cos^2A

[1-sin^2A = cos^2A]

= (1-sinA)/cos A hence proved

[ cosA and cosA cancels out)