Question

Prove that 1/sec theta-tan theta = 1+sin theta/cos theta

Answer 1

Answer:

by rhs

1+sin theta/cos theta

1/cos theta +sin theta

since 1/sec theta =sec theta

and sin theta /cos theta = tan theta

therefore 1/sec theta - tan theta

hence proved

Step-by-step explanation:

now pls mark me branliest

Answer 2

Step-by-step explanation:

To prove :  $\frac{1}{secA-tanA }$  = $\frac{(1+sinA)}{cosA}$

LHS = $\frac{1}{secA-tanA }$

       = $\frac{1}{secA-tanA } * \frac{secA+tanA}{secA+tanA }$

       = $\frac{secA+tanA}{sec^2A-tan^2A}$

       = secA+tanA

       = $\frac{1}{cosA} + \frac{sinA}{cosA}$

       = $\frac{(1+sinA)}{cosA}$

       = RHS

=> LHS = RHS

Hence proved