Question

1-cosA/1+cosA=secA-1/secA+1​

Answer 1

Question :-

$\dfrac{1 -cos\:A}{1 + cos\:A} =\dfrac{sec\:\:A -1}{sec\:\:A +1}$  

Solution :-

Here,

L.H.S. = $\dfrac{1 -cos\:A}{1 + cos\:A}$

R.H.S. =  $\dfrac{sec\:\:A -1}{sec\:\:A +1}$

Solving R.H.S.,

$\rightarrow \dfrac{sec\:\:A -1}{sec\:\:A +1}$

$\mapsto \tt{We\:\:know\:\:that}\:\:\dfrac{1}{cos \theta}=sec \theta$

Now,

$\rightarrow \dfrac{\dfrac{1}{cos\:A}-1}{\dfrac{1}{cos\:A}+1}\\\\\\ \rightarrow \dfrac{\dfrac{1-1\times cos\:A}{cos\:A}}{\dfrac{1+1\times cos\:A}{cos\:A}}\\\\\\ \rightarrow \dfrac{\dfrac{1-cos\:A}{cos\:A}}{\dfrac{1+cos\:A}{cos\:A}}\\\\\\ \star \:\:cos\:A\:\:gets\:\:cancel\\\\ \rightarrow \dfrac{1-cos\:A}{1+cos\:A} = L.H.S.\\\\\\ \underline{Hence,\:\:proved.}$

More Information :-

$\bullet\:\sf Trigonometric\:Values :\\\\\boxed{\begin{tabular}{c|c|c|c|c|c}Angle & 0 & 30 & 45 & 60 & 90\\\cline{1-6}Sin \theta & 0 & \dfrac{1}{2} & \dfrac{1}{\sqrt{2}} & \dfrac{\sqrt{3}}{2} & 1\\\cline{1-6}Cos \theta & 1 & \dfrac{\sqrt{3}}{2}& \dfrac{1}{\sqrt{2}}& \dfrac{1}{2}&0\\\cline{1-6}Tan \theta&0& \dfrac{1}{\sqrt{3}}&1&\sqrt{3}&Not D{e}fined\end{tabular}}$

$\bullet\:\:sin\:\theta=\dfrac{1}{cosec\:\theta},\\\\cosec\:\theta=\dfrac{1}{sin\:\theta}\\\\\\\bullet\:\:cos\:\theta=\dfrac{1}{sec\:\theta},\\\\sec\:\theta=\dfrac{1}{cos\:\theta}\\\\\\\bullet\:\:tan\:\theta=\dfrac{1}{cot\:\theta}=\dfrac{sin\:\theta}{cos\:\theta},\\\\cot\:\theta=\dfrac{1}{tan\:\theta}=\dfrac{cos\:\theta}{sin\:\theta}\\\\\\\bullet\:\:sin\:\theta=cos(90^{\circ}-\theta),\\\\cos\:\theta=sin(90^{\circ}-\theta)$

$\bullet\:\:tan\:\theta=cot(90^{\circ}-\theta),\\\\cot\:\theta=tan(90^{\circ}-\theta)\\\\\\\bullet\:\:sec\:\theta=cosec(90^{\circ}-\theta),\\\\cosec\:\theta=sec(90^{\circ}-\theta)$