Question

If cos A = k, then the value of the expression (sin A)(cos A)(tan A) is equivalent to:

Answer 1

Answer:

$(\sin A)(\cos A)(\tan A)=1-k^2$

Step-by-step explanation:

$\textsf{Trig identity}: \quad \tan A=\dfrac{\sin A}{\cos A}$

$\begin{aligned}\implies (\sin A)(\cos A)(\tan A)& =(\sin A)(\cos A)\dfrac{(\sin A)}{(\cos A)}\\\\& =\dfrac{(\sin A)(\cos A)(\sin A)}{(\cos A)}\\\\& =\sin^2 A\\\end{aligned}$

$\begin{aligned}\textsf{Trig identity}: \quad \sin^2 A + \cos^2 A &=1\\\implies \sin^2 A & =1-\cos^2 A\end{aligned}$

$\implies (\sin A)(\cos A)(\tan A)=1-\cos^2 A$

If $\cos A=k$ then:

$(\sin A)(\cos A)(\tan A)=1-k^2$

Answer 2

Answer:

1-k^2

Step-by-step explanation:

tan A= sin A/cos A

(sin A)(cos A)(sin A/cos A)=(sin A)^2

(sin A) ^2 + (cos A) ^2= 1

(sin A) ^2=1-(cos A) ^2

(sin A)^2= 1 - k^2