Question

prove that sin A tan A/1-cos A =1+secA


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Answer 1

Answer:

sinA=2sinA/2.cosA/2

1-cosA=2sin²A/2.

so sinA/1-cosA=cot(A/2)

So, sinA.tanA/1-cosA=cot(A/2).tanA

=cos(A/2)/sin(A/2).sinA/cosA

=cos(A/2)/sin(A/2).2sin(A/2).cos(A/2)/cosA

=2cos²(A/2)/cosA

=1+cosA / cosA. ( as cosA=2cos²(A/2) - 1 )

=1+secA

Answer 2

Correct Question :-

Prove that:

$\sf{\dfrac{sin \: A . tan \: A}{1 - cos \: A} = 1 + sec \: A}$

Answer :-

L.H.S = $\sf{\dfrac{sin \: A . tan \: A}{1 - cos \: A}}$

$\sf{\longrightarrow \: \dfrac{sin \: A . tan \: A}{1 - cos \: A}}$

$\sf{\longrightarrow \: \dfrac{sin \: A \times tan \: A}{1 - cos \: A}}$

$\sf{\longrightarrow \: \dfrac{sin \: A \times \dfrac{sin \: A}{cos \: A}}{1 - cos \: A}} \: \: \: \: {}_{.. \: .. \: ..} \: \: \bigg \{ \: tan \: A = \dfrac{sin \: A}{cos \: A} \bigg \}$

$\sf{\longrightarrow \: \dfrac{\dfrac{{sin}^{2} \: A}{cos \: A}}{1 - cos \: A}} \: \: \: \: {}_{.. \: .. \: ..} \: \: \{multiplying \: sin \: A \: and \: sin \: A \}$

$\sf{\longrightarrow \: \dfrac{{sin}^{2} \: A}{cos \: A(1 - cos \: A)}} \: \: \: \: {}_{.. \: .. \: ..} \: \: \{transposing \: cos \: A \: to \: denominator \}$

$\sf{\longrightarrow \: \dfrac{1 - {cos}^{2} \: A}{cos \: A(1 - cos \: A)}}$

$\sf{\longrightarrow \: \dfrac{(1 - cos \: A)(1 + cos \: A)}{cos \: A(1 - cos \: A)}}$

$\sf{\longrightarrow \: \dfrac{ \cancel{(1 - cos \: A)}(1 + cos \: A)}{cos \: A \cancel{(1 - cos \: A)}}}$

$\sf{\longrightarrow \: \dfrac{1 + cos \: A}{cos \: A}}$

$\sf{ \longrightarrow \: \dfrac{1}{cos} + \dfrac{cos \: A}{cos \: A}}$

$\sf{ \longrightarrow \: \dfrac{1}{cos} + \cancel{ \dfrac{cos \: A}{cos \: A}}}$

$\sf{\longrightarrow \: \dfrac{1}{cos \: A} + 1 \: \: \: \: {}_{.. \: .. \: ..} \: \bigg \{ \: \dfrac{1}{cos \: A} = sec \: A \bigg \}}$

$\sf{\longrightarrow \: sec \: A + 1 = R.H.S}$

Hence, Proved!

⠀⠀⠀⠀⠀ ______________________

THINGS TO REMEMBER :-

• $\sf{\dfrac{1}{cos \: A} = sec \: A}$

• $\sf{tan \: A = \dfrac{sin \: A}{cos \: A}}$

• $\sf{1 - {cos}^{2} \: A = {sin}^{2} \: A}$