If sin A-B-C-1- tan A-B-1-3 and SecA-C-2- then

The correct option is D We have A+B+C=90^o, A B=30^o, and A+C=60^o ⇒ B=30^o, A=60^o, C=0^o ...

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L'Hospital Rule to Remove Indeterminate Form

If sin A+B+C=...

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If sin(A+B+C)=1, tan(A-B)= $\frac{1}{\sqrt{3}}$ and Sec(A+C)=2, then

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None of the above

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\frac{t a n A + s e c A - 1}{t a n A - s e c A + 1} = \frac{1 + s i n A}{c o s A}

\frac{1 + s i n A}{1 - s i n A} = s e c A - t a n A

\frac{s i n A - 2 s i n^{2} A}{2 c o s^{3} A - c o s A} = t a n A

\sqrt{\frac{1 + sin A}{1 - sinA}} = secA + tan A

\frac{1 - s i n A}{1 + s i n A} = (s e c A - t a n A)^{2}

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