Question

solve this question
$1 + \cos(a ) \div 1 - \cos(a) = \tan({a}^{2} ) \div ( \sec(a) - 1)^{2}$
​

Answer 1

Step-by-step explanation:

To Prove : ${\sf{\ \ {\dfrac{1 + cos A}{1 - cos A}} = {\dfrac{tan^2 A}{(sec A - 1)^2}}}}$

____________________________

R.H.S. = ${\sf{\ \ {\dfrac{tan^2 A}{(sec A - 1)^2}}}}$

____________________________

${\boxed{\tt{\bigstar \ \ Identity \ : \ 1 + tan^2 A = sec^2 A}}}$

${\tt{From \ this, \ we \ get, \ [ tan^2 A = sec^2 A - 1}}$

____________________________

$\implies{\sf{ {\dfrac{sec^2 A - 1}{(sec A - 1)^2}}}}$

____________________________

• We can write this as :

$\implies{\sf{ {\dfrac{(sec A)^2 - (1)^2}{(sec A - 1)^2}}}}$

____________________________

${\boxed{\tt{\bigstar \ \ Identity \ : \ a^2 - b^2 = (a + b)(a - b)}}}$

${\tt{Here, \ a = sec A, \ b = 1}}$

____________________________

$\implies{\sf{ {\dfrac{(sec A + 1)(sec A - 1)}{(sec A - 1)^2}}}}$

____________________________

• Cancelling the common term.

$\implies{\sf{ {\dfrac{sec A + 1}{sec A - 1}}}}$

____________________________

${\boxed{\tt{\bigstar \ \ sec A = {\dfrac{1}{cos A}}}}}$

____________________________

$\implies{\sf{ {\dfrac{ {\dfrac{1}{cos A}} + 1}{ {\dfrac{1}{cos A}} - 1}}}}$

____________________________

$\implies{\sf{ {\dfrac{ {\dfrac{1 + cos A}{cos A}} }{ {\dfrac{1 - cos A}{cos A}} }}}}$

____________________________

$\implies{\sf{ {\dfrac{1 + cos A}{1 - cos A}}}}$

____________________________

= L.H.S.

Hence, proved !!

Answer 2

QuesTioN :

$\normalsize\sf\frac{1 - cosA}{1 + cosA} = \frac{Tan^2A}{(secA - 1)^2}$

AnswEr :

$\underline{\textbf{Taking \: Right \: hand \: side-}}$

$\normalsize\ : \implies\sf\frac{Tan^2A}{(secA - 1)^2}$

$\scriptsize\sf{\: \: \: \: \: \:( \therefore\ \: \red{Tan^2A = sec^2A - 1 }) }$

$\normalsize\ : \implies\sf\frac{sec^2A - 1}{(sec^2 - 1)}$

$\scriptsize\sf{\: \: \: \: \: \:( \therefore\ \: \red{Using \: algebric \: identities}) }$

$\normalsize\ : \implies\sf\frac{(secA + 1) \cancel{(secA - 1)}}{\cancel{(secA -1)} (secA -1)} \\ \\ \normalsize\ : \implies\sf\frac{secA + 1}{secA - 1 }$

$\scriptsize\sf{\: \: \: \: \: \:( \therefore\ \: \red{SecA = \frac{1}{CosA}}) }$

$\normalsize\ : \implies\sf\frac{\frac{1}{cosA} + 1}{\frac{1}{cosA} - 1} \\ \\ \normalsize\ : \implies\sf\frac{\frac{1 + cosA}{\cancel{cosA}}}{\frac{1 - cosA}{\cancel{cosA}}} \\ \\ \normalsize\ : \implies\sf\frac{1 + cosA}{1 - cosA} \\ \\ \normalsize\ : \implies\sf\ Right \: hand \: side \\ \\ \normalsize\ : \implies\sf\ L.H.S = R.H.S$

$\normalsize\ : \implies{\underline{\boxed{\sf \green{Hence \: prove \: !! }}}}$

$\rule{100}2$

Some Important related to it :

$\boxed{\begin{minipage}{6cm} Important Trigonometric identities :- \\ \\ \: \: 1)\sin^2\theta+\cos^2\theta=1 \\ \\ 2)\sin^2\theta= 1-\cos^2\theta \\ \\ 3)\cos^2\theta=1-\sin^2\theta \\ \\ 4)1+\cot^2\theta=\text{cosec}^2 \, \theta \\ \\5) \text{cosec}^2 \, \theta-\cot^2\theta =1 \\ \\ 6)\text{cosec}^2 \, \theta= 1+\cot^2\theta \\\ \\ 7)\sec^2\theta=1+\tan^2\theta \\ \\ 8)\sec^2\theta-\tan^2\thetha=1 \\ \\ 9)\tan^2\theta=\sec^2\theta-1 \end{minipage}}$