Answer:
[tex](\sin A)(\cos A)(\tan A)=1-k^2[/tex]
Step-by-step explanation:
[tex]\textsf{Trig identity}: \quad \tan A=\dfrac{\sin A}{\cos A}[/tex]
[tex]\begin{aligned}\implies (\sin A)(\cos A)(\tan A)& =(\sin A)(\cos A)\dfrac{(\sin A)}{(\cos A)}\\\\& =\dfrac{(\sin A)(\cos A)(\sin A)}{(\cos A)}\\\\& =\sin^2 A\\\end{aligned}[/tex]
[tex]\begin{aligned}\textsf{Trig identity}: \quad \sin^2 A + \cos^2 A &=1\\\implies \sin^2 A & =1-\cos^2 A\end{aligned}[/tex]
[tex]\implies (\sin A)(\cos A)(\tan A)=1-\cos^2 A[/tex]
If [tex]\cos A=k[/tex] then:
[tex](\sin A)(\cos A)(\tan A)=1-k^2[/tex]
