Question

sinQ-cosQ+1/sinQ+cosQ-1=1/secQ-tanQ

Answer 1

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Answer 2

Solution:

LHS = $\frac{sinQ-cosQ+1}{sinQ+cosQ-1}$

Divide numerator and denominator by cosQ , we get

= $\frac{tanQ-1+secQ}{tanQ+1-secQ}$

=$\frac{secQ+tanQ-1}{1+tanQ-secQ}$

/* By Trigonometric identity:

Sec²Q-tan²Q = 1 */

= $\frac{secQ+tanQ-(sec^{2}Q-tan^{2}Q)}{1+tanQ-secQ}$

= $\frac{secQ+tanQ-(secQ+tanQ)(secQ-tanQ)}{1+tanQ-secQ}$

= $\frac{(secQ+tanQ)[1-(secQ-tanQ)]}{1+tanQ-secQ}$

= $\frac{(secQ+tanQ)[1-secQ+tanQ)]}{1-secQ+tanQ}$

After cancellation, we get

= $(secQ+tanQ)$

Multiply numerator and denominator by (secQ-tanQ), we get

= $\frac{(secQ+tanQ)(secQ-tanQ)}{secQ-tanQ}$

= $\frac{sec^{2}Q-tan^{2}Q}{secQ-tanQ}$

= $\frac{1}{secQ-tanQ}$

= RHS