Question

tanA+ secA-1/tanA + secA+1=1+sinA/cosA

​

Answer 1

Correct Question:

Q. Prove that  $\sf{\dfrac{tanA+secA-1}{tanA-secA+1}=\dfrac{1+sinA}{cosA} }$.

Answer:

⟼To Prove:

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

❀Solution❀

We know that,

$\rightarrow\sf{1+tan^{2}x=sec^{2}x\implies1=sec^{2}x-tan^{2}x}$

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

$\rightarrow\sf{tanx=\dfrac{sinx}{cosx}}$

$\rightarrow\sf{secx=\dfrac{1}{cosx}}$

Now,

$\longrightarrow\sf{L.H.S.=\dfrac{tanA+secA-1}{tanA-secA+1}}$

$\longrightarrow\sf{L.H.S.=\dfrac{tanA+secA-(sec^{2}A-tan^{2}A)}{tanA-secA+1}}$

$\longrightarrow\sf{L.H.S.=\dfrac{secA+tanA-(secA+tan)(secA-tanA)}{tanA-secA+1}}$

$\longrightarrow\sf{L.H.S.=\dfrac{(secA+tanA)(1-(secA-tanA))}{tanA-secA+1}}$

$\longrightarrow\sf{L.H.S.=\dfrac{(secA+tanA)(1-secA+tanA)}{tanA-secA+1}}$

$\longrightarrow\sf{L.H.S.=\dfrac{(secA+tanA)(tanA-secA+1)}{tanA-secA+1}}$

$\longrightarrow\sf{L.H.S.=secA+tanA}$

$\longrightarrow\sf{L.H.S.=\dfrac{1}{cosA} +\dfrac{sinA}{cosA} }$

$\longrightarrow\sf{L.H.S.=\dfrac{1+sinA}{cosA}}$

$\longrightarrow\sf{L.H.S.=R.H.S.}$

Hence Proved

Answer 2

To Prove :

$\sf{\frac{ \tan(A) + \sec(A )- 1 }{ \tan(A) - \sec(A) + 1 }} = \frac{1 + \sin(A) }{ \cos(A) } \:$

Proof:

$\sf{ \:LHS \: = \frac{ \tan(A) \sec(A) - ( { \sec(A) }^{2} - { \tan(A) }^{2}) }{ \tan(A - \sec(A) + 1) } } \\ \\ \sf{LHS = \frac{ \sec(A) + \tan(A)( \sec(A) + \tan(A) ( \sec(A) - \tan(A) ) }{ \tan(A) - \sec(A) + 1 } } \\ \\ \sf{LHS \: = \frac{ \sec(A) + \tan(A)(1 - ( \sec(A) - \tan(A) ) }{ \tan(A - \sec(A) + 1) } } \\ \\ \sf{ \: LHS = \frac{ \sec(A) + \tan(A) ( \tan(A) - \sec(A) + 1)}{ \tan(A) - \sec(A) + 1 }} \\ \\ \sf{ LHS = \sec(A) + \tan(A) } \\ \\ \sf{LHS \: = \frac{1}{ \cos(A)}} + \frac{ \sin(A) }{ \cos(A) } \\ \\ \sf{LHS = \frac{1 + \sin(A) }{ \cos(A) }} \\ \\ \sf{LHS = RHS}$