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\( \frac{\tan x+\sec x-1}{\tan x-\sec x+1}=\frac{1+\sin x}{\cos x} \) |
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Answer» L.H.S. = \(\frac{tanx + sec x-1}{tan x-sec x+1}\) = \(\frac{2tan x}{2sec x-2}\) = \(\frac{tan x}{sec x-1}\) = \(\frac{sinx}{1-cos x}\) = \(\cfrac{2sin\frac x2cos\frac x2}{2sin^2\frac x2}\) = cot x/2 R.H.S. = \(\frac{1+sin x}{cos x}\) = \(\cfrac{(sin \frac x2+cos\frac x2)^2}{(cos\frac x2+sin \frac x2)(cos\frac x2-sin\frac x2)}\) = \(\frac{sin\frac x2+cos\frac x2}{cos\frac x2-sin\frac x2}\) = \(\frac{2cos\frac x2}{2sin\frac x2}=\) cot x/2 Hence, L.H.S. = R.H.S. ⇒ \(\frac{tanx + sec x-1}{tan x-sec x+1}\) = \(\frac{1+sinx}{cos x}\) Hence Proved |
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