Question

secA+tanA-1/tanA-secA+1=secA+tanA​

Answer 1

To prove:-

• $\sf \dfrac{tanA + secA - 1}{tanA - secA + 1} = secA + tanA$

Proof:-

★ Identity used:-

$:\implies$ sec²A - tan²A = 1

We have,

$\sf \dfrac{tanA + secA - 1}{tanA - secA + 1} = secA + tanA \\\\\\\small\sf\;\;\;\star\; \underline{Taking\;L.H.S:-}\\\\\\:\implies\sf \dfrac{tanA + secA - 1}{tanA - secA + 1}\\\\\\:\implies\sf \dfrac{tanA + secA - (sec^2A - tan^2A)}{tanA - secA + 1}\\\\\\:\implies\sf \dfrac{tanA + secA - ((secA + tanA)(secA - tanA))}{tanA - secA + 1}\\\\\\:\implies\sf \dfrac{(tanA + secA)(1 - (secA - tanA))}{tanA - secA + 1}\\\\\\:\implies\sf \dfrac{(tanA + secA) \cancel{(1 - secA + tanA)}}{ \cancel{tanA - secA + 1}}\\\\\\:\implies\sf tanA + secA\\\\\\:\implies\sf LHS = RHS$

$\dag$ Hence, ProvEd!

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✩ Fundamental Trigonometric identities

$\begin{lgathered}\boxed{\begin{minipage}{15 em} \sf \displaystyle \bullet \cos^{2}\theta + sin^{2}\theta=1 \\\\\\ \bullet \ 1 + tan^{2}\theta=\sec^{2}\theta \\\\\\ \bullet \ 1 + cot^{2}\theta = cosec^{2}\theta \end{minipage}}\end{lgathered}$