Respuesta :
I assume that means: Show
[tex]\dfrac{1}{\cot a + \tan A} = \sin A \cos A[/tex]
Proof:
[tex]\dfrac{1}{\cot A + \tan A}[/tex]
[tex]= \dfrac{1}{\frac{\cos A}{\sin A} + \frac{\sin A}{\cos A}}[/tex]
[tex]= \dfrac{1}{\frac{\cos^2A + \sin^2 A}{\sin A \cos A}}[/tex]
[tex]= \dfrac{\sin A \cos A}{\cos^2A + \sin^2 A}[/tex]
[tex]=\sin A \cos A \quad\checkmark[/tex]
ANSWER:
The given statement [tex]$\frac{1}{\cot A+\tan A}=\sin A \times \cos A$[/tex] has been proved
SOLUTION:
We need to prove that, [tex]$\frac{1}{\cot A+\tan A}=\sin A \times \cos A$[/tex]
Now, take left hand side, and by solving it we have bring up the right hand side.
[tex]L.H.S = \frac{1}{\cot A+\tan A}$[/tex]
Since, we know that, [tex]\tan \theta=\frac{\sin \theta}{\cos \theta}$ and $\cot \theta=\frac{\cos \theta}{\sin \theta}$[/tex]
Substitute in above L.H.S we get,
[tex]$=\frac{1}{\frac{\cos A}{\sin A}+\frac{\sin A}{\cos A}}$[/tex]
On cross- multiplication, the above expression becomes,
[tex]$=\frac{1}{\frac{\cos A \times \cos A+\sin A \times \sin A}{\sin A \times \cos A}}$[/tex]
On simplification we get,
[tex]$=\frac{1}{\frac{\sin A^{2}+\cos A^{2}}{\sin A \times \cos A}}$[/tex]
we know that the trignometric identity, [tex]$\sin \theta^{2}+\cos \theta^{2}=1$[/tex] the above equation becomes
[tex]$=\frac{1}{\frac{1}{\sin A \times \cos A}}$[/tex]
[tex]=\frac{1}{1} \times \frac{\sin A \times \cos A}{1}$=\sin A \times \cos A$[/tex]
= R.H.S
L.H.S = R.H.S
Hence, the given statement has been proved.
