Answer:
Step-by-step explanation:
Right side =
sin A / cos A + sinB/ cosB (sinAcosB + sinB cos A ) * cosA cosB
------------------------------------- = cosA cosB (sinAcosB - snBcosA ) sinA/cosA - sinB/cos B
= Left side.
The trigonometry identity [tex]\frac{sin(A+B)}{sin(A-B)}[/tex] is equals to [tex]\frac{tan(A)+tan(B)}{tan(A)-tan(B)}[/tex].
Trigonometric Identities are the equalities that involve trigonometry functions and holds true for all the values of variables given in the equation.
According to the given question.
We have a trigonometric identity.
[tex]\frac{sin(A+B)}{sin(A-B)} =\frac{tan(A)+tan(B)}{tan(A)-tan(B)}[/tex]
To prove the above trigonometric identity we will show L.H.S = R.H.S
[tex]L.H.S=\frac{sin(A+B)}{sin(A-B)}[/tex]
⇒ [tex]L.H.S = \frac{cosBsinA-sinBcosA}{sinAcosB-cosAsinB}[/tex]
⇒ [tex]L.H.S = \frac{\frac{sinAcosB}{cosAcosB} + \frac{sinBsinA}{cosBcosA} }{\frac{sinAcosB}{cosAcosB}-\frac{cosAsinB}{cosAcosB} }[/tex] (dividing the numerator and denominator by [tex]cosAcosB[/tex] )
⇒ [tex]L.H.S = \frac{\frac{sinA}{cosA} +\frac{sinB}{cosB} }{\frac{sinA}{cosA}-\frac{sinB}{cosB} }[/tex]
⇒ [tex]L.H.S = \frac{tanA+tanB}{tanA- tanB}= R.H.S[/tex]
Hence, L.H.S = R.H.S
Find out more information about trigonometric identities here:
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