(High point value) Last trig qn for today from me :
(1+cosA)/(1-cosA)=tan²A/(secA -1)²
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(qn no 27 in the photo but both are same method)
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sorrryy idk...........
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Answer:
\frac{tan^{2}A}{(secA-1)^{2}}=\frac{1+cosA}{1-cosA}
Step-by-step explanation:
LHS=\frac{tan^{2}A}{(secA-1)^{2}}
=\frac{(sec^{2}A-1)}{(secA-1)^{2}}\\=\frac{(secA+1)(secA-1)}{(secA-1)(secA-1)}\\=\frac{(secA+1)}{(secA-1)}\\=\frac{\frac{1}{cosA}+1}{\frac{1}{cosA}-1}\\=\frac{\frac{(1+cosA)}{cosA}}{\frac{(1-cosA)}{cosA}}\\=\frac{1+cosA}{1-cosA}\\=RHS
\frac{tan^{2}A}{(secA-1)^{2}}=\frac{1+cosA}{1-cosA}
Step-by-step explanation:
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