Question

if tanA+secA=l then prove that secA=l2+l/2l

Answer 1

Given: tan + sec = l
To Prove : sec = $\dfrac{l^2+l} {2}$

Proof:
tan+sec = l - - - i)

As we know,
1+tan² = sec²
sec² - tan²= 1
(sec+tan) (sec-tan) =1
sec-tan = $\dfrac{1} {tan + sec}$
sec-tan = $\frac{1} {l}$ - - - ii)

Adding i) & ii)
sec+tan=l
sec-tan=$\dfrac{1} {l}$

2sec = l+$\dfrac{1} {l}$
2sec=$\dfrac{l^2 +1} {l}$
sec = $\dfrac{l^2+1} {2l}$

Hence proved!!

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