Math, asked by sudhasudhakar10, 9 months ago

SinAtanA/1-cosA=1+secA

Answers

Answered by Anonymous

❥ ${\underline{\underline{\large{\mathtt{QUESTION:-}}}}}$

Prove that,

$\sf\frac{sinA\:tanA}{1-cosA}=1+secA$

❥ ${\underline{\underline{\large{\mathtt{SOLUTION:-}}}}}$

Taking L.H.S ,

$\sf\frac{sinA\:tanA}{1-cosA}$

★we know that tanA = sinA/cosA★

$\implies\sf\frac{sinA\:\frac{sinA}{cosA}}{1-cosA}$

$\implies\sf\frac{\frac{sin^2A}{cosA}}{1-cosA}$

★We know that 1-cos²A = sin²A★

$\implies\sf\frac{\frac{1-cos^2A}{cosA}}{1-cosA}$

★Apply a²-b²=(a+b)(a-b)★

$\implies\sf\frac{\frac{(1+cosA)(1-cosA)}{cosA}}{1-cosA}$

$\implies\sf\frac{(1+cosA)(1-cosA)}{cosA(1-cosA)}$

$\implies\sf\frac{1+cosA}{cosA}$

★Apply the rule of separation★

$\implies\sf\frac{1}{cosA}+\frac{cosA}{cosA}$

$\implies\sf{secA+1}$

$\implies\sf{1+secA\:(Proved)}$

L.H.S = R.H.S

Answered by MrBhukkad

$\huge{ \bigstar}\huge{\mathcal{ \overbrace{ \underbrace{ \pink{ \fbox{ \green{ \blue{S} \pink{o} \red{l} \green{u} \purple{t} \blue{i} \green{o} \red{n}}}}}}}} \huge{ \bigstar}$

$\tt{ \underline{ \underline{ \red{Question}}}}: - \\ \bf{Prove \: that,} \\ \bf{ \frac{sinAtanA}{1 - cosA} = 1 + secA} \\ \tt{ \underline{ \underline{ \orange{Answer}}}}: -$

Taking L.H.S.,

$\\ \\ \bf{Formulas \: used} \\ \boxed{ \bf{ 1 - {cos}^{2}A = {sin}^{2}A}} \\ \boxed{ \bf{ {a}^{2} - {b}^{2} = (a + b)(a - b) + }} \\ \\ \longrightarrow \bf {\frac{sinAtanA}{1 - cosA} } \\ \longrightarrow \bf{ \frac{sinA \frac{sinA}{cosA} }{1 - cosA} } \\ \longrightarrow \bf{ \frac{ \frac{ {sin}^{2}A }{cosA} }{1 - cosA} } \\ \longrightarrow \bf{ \frac{ \frac{1 - {cos}^{2} A}{cosA} }{1 - cosA} } \\ \longrightarrow \bf{ \frac{ \frac{(1 + cosA)(1 - cosA)}{cosA} }{1 - cosA} } \\ \longrightarrow \bf{ \frac{(1 + cosA) \cancel{(1 - cosA)}}{cosA \cancel{(1 - cosA)}} } \\ \longrightarrow \bf{ \frac{1}{cosA} + \frac{cosA}{cosA} } \\ \longrightarrow \bf{secA + 1} \\ \longrightarrow \bf{1 + secA} \: (Proved)$

L.HS = R.H.S.

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