Question

tan A + sec A -1/ tanA -sec A +1 = 1+ sinA/ cos A​

Answer 1

$\large{\underline{\underline{\boxed{\boxed{\mathfrak{\red{Question:}}}}}}}$

$\implies{\sf{\dfrac{tan\;A + \sec\;A - 1}{tan\;A - \sec\;A + 1}=\dfrac{1 + \sin\;A}{\cos\;A}}}$

$\large{\underline{\underline{\boxed{\boxed{\mathfrak{\red{Step\: By\; Step\; Explanation:}}}}}}}$

${\bf{\star{\underline{\blue{To\;Prove:}}}}}$

$\implies{\sf{\dfrac{tan\;A + \sec\;A - 1}{tan\;A - \sec\;A + 1}=\dfrac{1 + \sin\;A}{\cos\;A}}}$

Now, we will take LHS part,

$\implies{\sf{\dfrac{tan\;A + \sec\;A - 1}{tan\;A - \sec\;A + 1}}}$

$\implies{\sf{\dfrac{tan\;A + \sec\;A - (\sec^2\;A - \tan^2\; A)}{tan\;A - \sec\;A + 1}}}$

$\implies{\sf{\dfrac{tan\;A + \sec\;A - (\sec\;A + \tan\; A)(\sec\;A - \tan\; A)}{tan\;A - \sec\;A + 1}}}$

$\implies{\sf{\dfrac{(\tan\; A + \sec\; A)(-\sec\;A + \tan\;A + 1)}{tan\;A - \sec\;A + 1}}}$

$\implies{\sf{\tan\; A + \sec\; A}}$

$\implies{\sf{\dfrac{\sin\;A}{\cos\;A}+\dfrac{1}{\cos\;A}}}$

$\implies{\sf{\dfrac{\sin\; A + 1}{\cos\;A}}}$

∴ LHS = RHS

Hence Proved!!

Answer 2

Step-by-step explanation:

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