Question

$1 + \cos(a ) \div \sin(a) = \sin(a) \div 1 - \cos(a)$
prove that​

Answer 1

$\huge\tt{\red{\underline{Given:}}}$

★$\dfrac{1+cosA}{sinA}=\dfrac{sinA}{1-cosA}$

$\huge\tt{\red{\underline{To\:\:Prove:}}}$

★LHS=RHS.

$\huge\tt{\red{\underline{Concept\:\:Used:}}}$

★We would use some basic formulas related to trigonometry.

$\huge\tt{\red{\underline{Answer:}}}$

Taking RHS,

=$\dfrac{sinA}{1-cosA}$

=$\dfrac{sinA \times sinA}{(1-cosA) (sinA)}$

=$\dfrac{sin^{2}A}{(sinA) (1-cosA) }$

$\large\purple{\boxed{sin^{2}\theta+cos^{2}\theta = 1}}$

=$\dfrac{(1-cos^{2}A)}{(sinA) (1-cosA) }$

=$\dfrac{(1+cosA) (1-cosA) }{(sinA) (1-cosA)}$

$\large\green{\boxed{a^{2}-b^{2}=(a+b) (a-b) }}$

=$\dfrac{(1+cosA) \cancel{(1-cosA)} }{(sinA) \cancel{(1-cosA)}}$

=$\dfrac{1+cosA}{sinA}$

=RHS.

${\underline{\boxed{.°. LHS=RHS}}}$

Hence proved.