Question

prove that:
(1+cotA-cosecA)(1+tanA+secA)=2

Answer 1

Convert the left side in terms of sines and cosines to get:

LHS = [1 + cot(A) - csc(A)][1 + tan(A) + sec(A)]

= [1 + cos(A)/sin(A) - 1/sin(A)][1 + sin(A)/cos(A) + 1/cos(A)]

= [sin(A) + cos(A) - 1]/sin(A) * [sin(A) + cos(A) + 1]/cos(A), by getting common denominators

= {[sin(A) + cos(A)]^2 - 1}/[sin(A)cos(A)], via difference of squares

= [sin^2(A) + 2sin(A)cos(A) + cos^2(A) - 1]/[sin(A)cos(A)], by expanding

= [2sin(A)cos(A) + 1 - 1]/[sin(A)cos(A)], since sin^2(A) + cos^2(A) = 1

= [2sin(A)cos(A)]/[sin(A)cos(A)]

= 2, by canceling sin(A)cos(A)

= RHS.

I hope this helps!

Answer 2

$(1 + cota - coseca)(1 + tana + cota) \\ \\ = ( \frac{sina + cosa - 1}{sina} )( \frac{cosa + sina + 1}{cosa} ) \\ \\ = \frac{( {sina + cosa)}^{2} - 1 }{sinacosa} \\ \\ = \frac{2sinacosa}{sinacosa} \\ \\ = 2$