Question

Prove the following identities:
@ganeshie8

Answer 1

\[\frac{ \sin~2A+\cos~2A-1 }{ \sin~2A+\cos~2A+1 }=\frac{ \tan~A }{ \tan(45+A) }\]

Answer 2

we must get rid of the 1,right?

Answer 3

Not really sure
you want to replace 1 by sin^2A + cos^2A ?

Answer 4

no,i want to replace with the DA formulae..
so,can we use the DA formulae to replace 1?

Answer 5

Oh that's a nice idea

Answer 6

replace cos(2A) in top by 1 - 2sin^2A
replace cos(2A) in bottom by 2cos^2A - 1

?

Answer 7

yep..,that's what i'm thinking..but once i did it i got stuck somewhere.. xD

Answer 8

\[\begin{align}
\dfrac{ \sin~2A+\color{red}{\cos~2A}-1 }{ \sin~2A+\color{red}{\cos~2A}+1 }
&= \dfrac{ \sin~2A+\color{red}{1-2\sin^2A}-1 }{ \sin~2A+\color{red}{2\cos^2A-1}+1 }\\~\\

\end{align}\]

Answer 9

\[\begin{align}

\dfrac{ \sin~2A+\color{red}{\cos~2A}-1 }{ \sin~2A+\color{red}{\cos~2A}+1 }

&= \dfrac{ \sin~2A+\color{red}{1-2\sin^2A}-1 }{ \sin~2A+\color{red}{2\cos^2A-1}+1 }\\~\\

&= \dfrac{ \sin~2A\color{red}{-2\sin^2A} }{ \sin~2A+\color{red}{2\cos^2A} }\\~\\

\end{align}\]

Answer 10

Perhaps replace sin2A by 2sinAcosA and try factoring

Answer 11

okay

Answer 12

\[\begin{align}

\dfrac{ \sin~2A+\color{red}{\cos~2A}-1 }{ \sin~2A+\color{red}{\cos~2A}+1 }

&= \dfrac{ \sin~2A+\color{red}{1-2\sin^2A}-1 }{ \sin~2A+\color{red}{2\cos^2A-1}+1 }\\~\\

&= \dfrac{ \sin~2A\color{red}{-2\sin^2A} }{ \sin~2A+\color{red}{2\cos^2A} }\\~\\


&= \dfrac{ 2\sin A\cos A\color{red}{-2\sin^2A} }{ 2\sin A\cos A+\color{red}{2\cos^2A} }\\~\\


\end{align}\]

Answer 13

we can factorise the 2

Answer 14

2 cancels out

Answer 15

pull out sinA from top
pull out cosA from bottom

Answer 16

that gives you tanA

Answer 17

\[\begin{align}

\dfrac{ \sin~2A+\color{red}{\cos~2A}-1 }{ \sin~2A+\color{red}{\cos~2A}+1 }

&= \dfrac{ \sin~2A+\color{red}{1-2\sin^2A}-1 }{ \sin~2A+\color{red}{2\cos^2A-1}+1 }\\~\\

&= \dfrac{ \sin~2A\color{red}{-2\sin^2A} }{ \sin~2A+\color{red}{2\cos^2A} }\\~\\


&= \dfrac{ 2\sin A\cos A\color{red}{-2\sin^2A} }{ 2\sin A\cos A+\color{red}{2\cos^2A} }\\~\\

&= \dfrac{ \sin A(\cos A\color{red}{-\sin A)} }{ \cos A(\sin A+\color{red}{\cos A)} }\\~\\

\end{align}\]

Answer 18

Yep,i got stuck here..

Answer 19

\[\begin{align}

\dfrac{ \sin~2A+\color{red}{\cos~2A}-1 }{ \sin~2A+\color{red}{\cos~2A}+1 }

&= \dfrac{ \sin~2A+\color{red}{1-2\sin^2A}-1 }{ \sin~2A+\color{red}{2\cos^2A-1}+1 }\\~\\

&= \dfrac{ \sin~2A\color{red}{-2\sin^2A} }{ \sin~2A+\color{red}{2\cos^2A} }\\~\\


&= \dfrac{ 2\sin A\cos A\color{red}{-2\sin^2A} }{ 2\sin A\cos A+\color{red}{2\cos^2A} }\\~\\

&= \dfrac{ \sin A(\cos A\color{red}{-\sin A)} }{ \cos A(\sin A+\color{red}{\cos A)} }\\~\\

&= \dfrac{ \tan A(\cos A\color{red}{-\sin A)} }{ \sin A+\color{red}{\cos A} }\\~\\


\end{align}\]

Answer 20

maybe divide cosA top and bottom

Answer 21

alrite

Answer 22

\[\begin{align}

\dfrac{ \sin~2A+\color{red}{\cos~2A}-1 }{ \sin~2A+\color{red}{\cos~2A}+1 }

&= \dfrac{ \sin~2A+\color{red}{1-2\sin^2A}-1 }{ \sin~2A+\color{red}{2\cos^2A-1}+1 }\\~\\

&= \dfrac{ \sin~2A\color{red}{-2\sin^2A} }{ \sin~2A+\color{red}{2\cos^2A} }\\~\\


&= \dfrac{ 2\sin A\cos A\color{red}{-2\sin^2A} }{ 2\sin A\cos A+\color{red}{2\cos^2A} }\\~\\

&= \dfrac{ \sin A(\cos A\color{red}{-\sin A)} }{ \cos A(\sin A+\color{red}{\cos A)} }\\~\\

&= \dfrac{ \tan A(\cos A\color{red}{-\sin A)} }{ \sin A+\color{red}{\cos A} }\\~\\


&= \tan A *\dfrac{(\cos A\color{red}{-\sin A)}/\color{blue}{\cos A} }{ (\sin A+\color{red}{\cos A})/\color{blue}{\cos A} }\\~\\


\end{align}\]

Answer 23

\[\begin{align}

\dfrac{ \sin~2A+\color{red}{\cos~2A}-1 }{ \sin~2A+\color{red}{\cos~2A}+1 }

&= \dfrac{ \sin~2A+\color{red}{1-2\sin^2A}-1 }{ \sin~2A+\color{red}{2\cos^2A-1}+1 }\\~\\

&= \dfrac{ \sin~2A\color{red}{-2\sin^2A} }{ \sin~2A+\color{red}{2\cos^2A} }\\~\\


&= \dfrac{ 2\sin A\cos A\color{red}{-2\sin^2A} }{ 2\sin A\cos A+\color{red}{2\cos^2A} }\\~\\

&= \dfrac{ \sin A(\cos A\color{red}{-\sin A)} }{ \cos A(\sin A+\color{red}{\cos A)} }\\~\\

&= \dfrac{ \tan A(\cos A\color{red}{-\sin A)} }{ \sin A+\color{red}{\cos A} }\\~\\


&= \tan A *\dfrac{(\cos A\color{red}{-\sin A)}/\color{blue}{\cos A} }{ (\sin A+\color{red}{\cos A})/\color{blue}{\cos A} }\\~\\

&=\tan A * \dfrac{1-\tan A}{\tan A + 1}

\end{align}\]

Answer 24

next replace 1 by tan(45)

Answer 25

nice :)

Answer 26

ook

Answer 27

\[tanA~\times~\frac{ \tan45-tanA }{tanA-\tan45 }\]

Answer 28

\[tanA~\times~\frac{ \tan45-tanA }{tanA\color{red}{+}\tan45 }\]

Answer 29

lookup tan(A+B) formula

Answer 30

tan(A+B)=(tanA+tanB)/(1-tanAtanB)

Answer 31

Oh one sec

Answer 32

\[\begin{align}

\dfrac{ \sin~2A+\color{red}{\cos~2A}-1 }{ \sin~2A+\color{red}{\cos~2A}+1 }

&= \dfrac{ \sin~2A+\color{red}{1-2\sin^2A}-1 }{ \sin~2A+\color{red}{2\cos^2A-1}+1 }\\~\\

&= \dfrac{ \sin~2A\color{red}{-2\sin^2A} }{ \sin~2A+\color{red}{2\cos^2A} }\\~\\


&= \dfrac{ 2\sin A\cos A\color{red}{-2\sin^2A} }{ 2\sin A\cos A+\color{red}{2\cos^2A} }\\~\\

&= \dfrac{ \sin A(\cos A\color{red}{-\sin A)} }{ \cos A(\sin A+\color{red}{\cos A)} }\\~\\

&= \dfrac{ \tan A(\cos A\color{red}{-\sin A)} }{ \sin A+\color{red}{\cos A} }\\~\\


&= \tan A *\dfrac{(\cos A\color{red}{-\sin A)}/\color{blue}{\cos A} }{ (\sin A+\color{red}{\cos A})/\color{blue}{\cos A} }\\~\\

&=\tan A * \dfrac{1-\tan A}{\tan A + 1}\\~\\

&=\tan A * \dfrac{1-\color{purple}{\tan(45)}\tan A}{\tan A + \color{purple}{\tan(45)}}


\end{align}\]

Answer 33

that should work, see

Answer 34

Ok

Answer 35

\[=\tan~\div \frac{ tanA+\tan~45 }{ 1-\tan45(tanA) }\]

Answer 36

=tan A/ (tan(45+A)

Answer 37

Thank you @ganeshie8 ^-^