Question

Prove that :secA+tenA=cosA/sinA

Answer 1

Answer:

your answer is here buddy....

secA + tanA =LHS

1/cosA + sinA/cosA.....

... .(secA=1/cotA & tanA=sinAcosA)

now, cross multiply-

cosA + sinA

cos²A

cosA + sinA

cos²A cos²A

now taking roots for the denominator-

cosA + sinA

√cos²A √cos²A

then,

cosA + sinA

cosA cosA

(cosA/cosA get canceled)

SinA/cosA

(by reciprocal)

CosA = RHS

SinA

hence proved"

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Answer 2

Question.

$secx + tanx = \frac{cosx}{1 - sinx}$

Step-by-step explanation:

$consider \: rhs \\ \\ \frac{cosx}{1 - sinx} \\ \\ multiply \: \: with \: \: (1 + sinx) \\ \\ = \frac{cosx}{(1 - sinx)} \times \frac{(1 + sinx)}{(1 + sinx)} \\ \\ = \frac{cosx(1 + sinx)}{1 - {sin}^{2}x } \\ \\ = \frac{cosx(1 + sinx)}{ {cos}^{2} x} \\ \\ = \frac{1 + sinx}{cosx} \\ \\ = \frac{1}{cosx} + \frac{sinx}{cosx} \\ \\ = secx + tanx$

$formulae \\ \\ {cos}^{2} x + {sin}^{2} x = 1 \\ \\ cosx \: secx \: = 1 \\ \\ \frac{sinx}{cosx} = tanx$

HOPE IT HELPS