Question

(secA ­- tan
A)/(cosecA + cot

Answer 1

Answer:

we know that

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

we know that

cosec^{2}A-cot^{A}=1\\\\(cosecA+cotA)(cosecA-cotA)=1\\\\cosecA+cotA=\frac{1}{cosecA-cotA}....(2)

Now,\\\frac{secA+tanA}{cosecA+cotA}

by using equations (1)&(2)

=\frac{\frac{1}{secA-tanA}}{\frac{1}{cosecA-cotA}}

=(\frac{1}{secA-tanA}).(\frac{cosecA-cotA}{1})\\\\=\frac{cosecA-cotA}{secA-tanA}

I hope this answer helps you

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