Question

Plzzz solve the problem​

Answer 1

Given :-

$\rm \dfrac{tan A-Sin A }{tan A + Sin A } = \dfrac{Sec A-1}{Sec A + 1}$

To prove :

LHS = RHS

Proof :-

$\rm RHS = \dfrac{Sec A-1}{Sec A + 1}$

$\rm \:LHS = \dfrac{tan A-Sin A }{tan A + Sin A }$

$\rm \rightarrow\dfrac{tan A-Sin A }{tan A + Sin A }$

$\rm \rightarrow\dfrac{ \dfrac{Sin \: A}{cos \: A} -\dfrac{Sin \: A}{ 1} }{ \dfrac{Sin \: A}{cos \: A} + \dfrac{Sin \: A}{ 1}} \\ \\ \rm \rightarrow \: \dfrac{Sin \: A}{Sin \: A} \times \bigg(\dfrac{ \dfrac{1}{cos \: A} -1 }{ \dfrac{1}{cos \: A} +1}\bigg ) \\ \\ \rm \rightarrow \: 1 \times \dfrac{ secA -1 }{ secA +1}\\ \\ \rm \rightarrow \: \dfrac{ secA -1 }{ secA +1}$

Therefore, LHS = RHS

• hence proved!

Additional Information :-

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

Answer 2

Proper question :

Prove that :

$\sf \dfrac {tanA \ - \ SinA}{tanA \ + \ SinA} \ = \ \dfrac {SecA \ - \ 1}{SecA \ + \ 1}$

Given :

$\sf \dfrac {tanA \ - \ SinA}{tanA \ + \ SinA} \ = \ \dfrac {SecA \ - \ 1}{SecA \ + \ 1}$

To prove :

LHS = RHS

Proof :

$\quad \sf LHS \ = \ \dfrac {tanA \ - \ SinA}{tanA \ + \ SinA}$

$\quad \sf RHS \ = \ \dfrac {SecA \ - \ 1}{SecA \ + \ 1}$

$\longrightarrow \sf \dfrac {tanA \ - \ SinA}{tanA \ + \ SinA}$

$\longrightarrow \sf \dfrac {\dfrac {sinA}{cosA} \ - \ \dfrac {SinA}{1}}{\dfrac {SinA}{cosA} \ + \ \dfrac {SinA}{1}}$

$\longrightarrow \sf \dfrac {SinA}{SinA} \times \bigg( \dfrac {\dfrac {1}{cosA} \ - \ 1}{\dfrac {1}{cosA} \ + \ 1 } \bigg)$

$\longrightarrow \sf 1 \times \dfrac {secA \ - \ 1}{secA \ + \ 1}$

$\longrightarrow \sf \dfrac {secA \ - \ 1}{secA \ + \ 1}$

$\sf = \ RHS$

$\therefore \large \sf LHS \ = \ RHS$

Hence proved ✔︎