Prove the identity (sin theta + 1- cos theta) /(cos theta -1+sin theta) =( 1+sin theta) /cos theta
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First, we can consider
sinx−cosx+1/sinx+cosx−1
=(sinx−cosx+1)(sinx+cosx+1)/(sinx+cosx−1)(sinx+cosx+1)
Thus, the numerator can be written as
= (sinx−cosx)(sinx+cosx)+sinx−cosx+sinx+cosx+1
= (sinx)2−(cosx)2+2sinx+1(sinx)2−(cosx)2+2sinx+1
= 2(sinx)2+2sinx=2sinx(1+sinx)2(sinx)2+2sinx=2sinx(1+sinx) ----------- (I)
Whereas, the denominator can be written as
= (sinx+cosx)2−1(sinx+cosx)2−1
= (sinx)2+2sinxcosx+(cosx)2−1=2sinxcosx(sinx)2+2sinxcosx+(cosx)2−1=2sinxcosx ---------- (II)
Hence, (I) / (II) = (1 + sin x)/(cos x) (thus the trigonometric identity is proven).
I hope that helps!
sinx−cosx+1/sinx+cosx−1
=(sinx−cosx+1)(sinx+cosx+1)/(sinx+cosx−1)(sinx+cosx+1)
Thus, the numerator can be written as
= (sinx−cosx)(sinx+cosx)+sinx−cosx+sinx+cosx+1
= (sinx)2−(cosx)2+2sinx+1(sinx)2−(cosx)2+2sinx+1
= 2(sinx)2+2sinx=2sinx(1+sinx)2(sinx)2+2sinx=2sinx(1+sinx) ----------- (I)
Whereas, the denominator can be written as
= (sinx+cosx)2−1(sinx+cosx)2−1
= (sinx)2+2sinxcosx+(cosx)2−1=2sinxcosx(sinx)2+2sinxcosx+(cosx)2−1=2sinxcosx ---------- (II)
Hence, (I) / (II) = (1 + sin x)/(cos x) (thus the trigonometric identity is proven).
I hope that helps!
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