Math, asked by aayushjoshi770, 3 months ago

Sin A -Cos A+1 / Sin A + Cos A - 1 = 1/ Sec A -Tan A​

Answers

Answered by ravi2303kumar
1

Answer:

to prove:

\frac{sinA-cosA+1}{sinA+cosA-1} = \frac{1}{secA-tanA}

take LHS,

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

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

=  \frac{tanA-1+secA}{tanA+1-secA}

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

= \frac{tan^2A-sec^2A -(tanA-secA)}{(tanA-secA +1)(tanA-secA)}

= \frac{(-1-tanA+secA)}{(tanA-secA +1)(tanA-secA)}

= \frac{-(tanA-secA +1)}{(tanA-secA +1)(tanA-secA)}

= \frac{-1}{(tanA-secA)}

= \frac{1}{(secA-tanA)}

= RHS

=> LHS = RHS

Hence proved

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