Question

Prove that sinA tanA÷1-cos A=1+secA

Answer 1

Step-by-step explanation:

Given :

$\rightarrowtail \bold{\dfrac{sinA\;tanA}{1-cosA} }$

To prove :

$\rightarrowtail \bold{\dfrac{sinA\;tanA}{1-cosA} =1+secA}$

Solution :

$\mapsto \bold{\dfrac{sinA\;tanA}{1-cosA}}$

• Rationalizing the denominator.

$\mapsto \bold{\dfrac{sinA\;tanA}{1-cosA}\times\dfrac{1+cosA}{1+cosA} }$

• ∵ (a - b)(a + b) = a² - b²

$\mapsto \bold{\dfrac{(sinA\;tanA)(1+cosA)}{(1)^{2}-(cos^{2})} }$

• ∵ tanA = sinA/cosA
• ∵ sin²A + cos²A = 1
• ∴ 1 - cos²A = sin²A

$\mapsto \bold{\dfrac{sinA(\dfrac{sinA}{cosA})(1+cosA) }{sin^{2}A} }$

$\mapsto \bold{\dfrac{\dfrac{sin^{2}A(1+cosA)}{cosA} }{sin^{2}A} }$

• sin²A, sin²A get cancelled.

$\Rightarrow \bold{\dfrac{1+cosA}{cosA} }$

$\Rightarrow \bold{\dfrac{1}{cosA}+\dfrac{cosA}{cosA} }$

• ∵ 1/cosA = secA

$\Rightarrow \bold{secA+1}$

$\twoheadrightarrow \bold{1+secA}$

• ∴ L.H.S = R.H.S