Question

tan theta+sec theta-1/ tan theta - sec theta + 1 = 1+sin theta/cos theta please be fast!!​

Answer 1

$\sf{Multiply \: and \: divide \: by \: tanA + secA - 1}$

$\implies \sf{ \frac{tanA +( secA - 1)}{ tanA -(secA - 1)} \times \frac{tanA +( secA - 1)}{tanA +( secA - 1)}} \\ \\ \implies \sf{ \frac{(tanA +( secA - 1)){}^{2} }{(tanA - ( secA - 1))(tanA +( secA - 1))} } \\ \\ \implies \sf{ \frac{(tanA +( secA - 1)) {}^{2} }{tan {}^{2} A - (sec A - 1){}^{2} } } \\ \\ \implies \sf{ \frac{tan {}^{2} A +( sec {}^{2}A + 1 - 2secA ) + 2tanA(secA - 1)}{tan {}^{2}A - (sec {}^{2}A + 1 -2 secA) } } \\ \\ \implies\sf{ \frac{tan {}^{2} A +1 +sec {}^{2} A- 2secA + 2tanAsecA -2tanA}{tan {}^{2}A - sec {}^{2}A - 1 + 2 secA } } \\ \\ \implies \sf{ \frac{ sec {}^{2}A +sec {}^{2}A - 2secA + 2sec AtanA - 2tanA}{ - 1 - 1 + 2secA }} \\ \\ \implies \sf{ \frac{ sec {}^{2}A - sec + sec AtanA - tanA}{secA - 1}} \\ \\ \implies \sf{ \frac{ secA(secA - 1) + tan A(secA - 1)\: }{secA - 1}} \\ \\ \sf{ \implies \frac{(secA - 1)(secA + tanA)}{(secA - 1)} } \\ \\ \implies \sf{secA + tanA} \\ \\ \implies \sf{ \frac{1}{cosA} + \frac{sinA}{cosA} \: } \\ \\ \implies \sf{ \frac{1 + sinA}{cosA}}$

$\sf{Using:}$

$\sf{line3: \: (a+b)(a-b)=a^2-b^2}$

$\sf{line4: \: (a+b)^2=a^2+b^2+2ab \: for\:numerator}$

$\sf{line5: \: tan^2A + 1 = sec^2 A}$

$\sf{line6: \: tan^2A - sec^2A = - 1}$

$\sf{line7: \: taking \:common (2), cancelling\: it}$

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