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# I need help proving the following trigonometric function:?

(secA + 1 )/(sin A + tan A) = (1)/(sinA)

I have to show all steps and only alter one side to make it equal to the other and I cannot do this problem. Help would be very appreciated thank you

### 3 Answers

- Anonymous9 years agoFavorite Answer
* LHS *

Note that tan(A) = sin(A)/cos(A) and sec(A) = 1/cos(A). Then..

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

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

Factoring the bottom expression and reducing the common factors of 1 + cos(A), we have..

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

= 1/sin(A)

= RHS

I hope this helps!

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- vohsLv 44 years ago
Wow you're rather on a roll. Asking a number of those questions approximately trigonometry. besides this one is trouble-free. Mulitply the denominator and the numerator contained in the best hand element with (a million + Cos B). The denominator will become ( a million - Cos^2 B) = Sin^2 B. Now merely divide the two Sin^3 B by utilising Sin ^2 B and multiply it with ( a million + Cos B).

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- MoonRoseLv 79 years ago
(secA + 1)/(sinA + tanA) (write secant and tangent in terms of sine and cosine)

(1/cosA + 1)/(sinA + sinA/cosA) (multiply the numerator and denominator by cosA)

(1 + cosA)/(cosAsinA + sinA) (factor out sinA from the denominator)

(1 + cosA) / sinA(cosA + 1) (cancel out 1 + cosA from the numerator and denominator)

1/sinA

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