Question

tana_seca+1/tana+seca-1​

Answer 1

Answer:

Step-by-step explanation:

Given,

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

Now we know,

sec²A - tan²A = 1

Substituting this identity,

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

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

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

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

= secA+tanA

Hence proved

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