Question

If secQ - tanQ =4, then prove that cosQ=8/17

Answer 1

We will start off by using the Trigonometric Equation given. We will try to form a quadratic equation in a suitable variable and work from there.

$\displaystyle \sec Q - \tan Q=4 \\ \\ \\ \implies \frac{1}{\cos Q}-\frac{\sin Q}{\cos Q}=4\\\\\\ \implies \frac{1-\sin Q}{\cos Q}=4\\\\\\\implies 1-\sin Q=4\cos Q \\\\\\ \textsf{We aim to form a quadratic equation. Squaring both sides} \\\\\\ \implies (1-\sin Q)^2=(4\cos Q)^2 \\\\\\ \implies 1-2\sin Q+\sin^2 Q=16\cos^2 Q \\\\\\ \left[\textsf{Use }\cos^2\theta = 1-\sin^2\theta \right] \\ \\ \\ \implies 1-2\sin Q+\sin^2 Q=16(1-\sin^2Q)\\\\\\\implies 1-2\sin Q+\sin^2Q=16-16\sin^2Q$

$\displaystyle \implies 17\sin^2Q-2\sin Q-15=0 \\\\\\ \implies 17\sin^2 Q-17\sin Q+15\sin Q-15=0 \\\\\\ \implies 17\sin Q(\sin Q-1)+15(\sin Q-1)=0\\\\\\ \implies (\sin Q-1)(17\sin Q+15)=0 \\\\\\ \implies \sin Q-1=0 \quad \text{OR} \quad 17\sin Q+15=0 \\\\\\ \implies \sin Q=1 \quad \text{OR} \quad \sin Q= -\frac{15}{17}$

But $\sin Q=1$ is not a valid solution, because then $\cos Q=0$ and $\sec Q-\tan Q \neq 4$. The question statement must remain valid.

Hence, $\sin Q=1$ cannot be a solution.

So,

$\displaystyle \sin Q = -\frac{15}{17} \\ \\ \\ \textsf{Use the identity: } \cos^2 \theta = 1-\sin^2\theta \\ \\ \\ \cos^2 Q=1-\left(-\frac{15}{17}\right)^2 \\\\\\ \implies \cos^2 Q = 1-\frac{225}{289} \\\\\\ \implies \cos^2 Q=\frac{289-225}{289}\\\\\\\implies \cos^2 Q=\frac{64}{289} \\\\\\ \implies \cos Q = \pm \frac{8}{17}$

Now, $\sin Q$ was negative. So, Q must be either in the Third Quadrant or the Fourth Quadrant.

If we take the third quadrant, then $\cos Q = -\dfrac{8}{17}$. But in this case,

$\displaystyle \begin{aligned}\sec Q-\tan Q &= -\frac{17}{8}-\left(\frac{-\frac{15}{17}}{-\frac{8}{17}}\right) \\ \\ \\ &= -\frac{17}{8}-\frac{15}{8} \\ \\ \\ &= \frac{-32}{8} \\ \\ \\ &= -4 \end{aligned}$

Which is not valid. So, Q cannot be in the third quadrant.

Hence, Q must be in the 4th quadrant, where $\cos Q$ is positive.

$\implies \Large \boxed{\boxed{\sf cos\ Q=\frac{8}{17}}} \\ \\ \\ \mathcal{HENCE \quad PROVED}$

Answer 2

This question can be solved by forming the given equation into a quadratic equation.

So, here we go:-

$\displaystyle \sec Q - \tan Q=4 \\ \\ \\ \implies \frac{1}{\cos Q}-\frac{\sin Q}{\cos Q}=4\\\\\\ \implies \frac{1-\sin Q}{\cos Q}=4\\\\\\\implies 1-\sin Q=4\cos Q \\\\\\ \textsf{Our goal is to form a quadratic equation. So, squaring both sides:-} \\\\\\ \implies (1-\sin Q)^2=(4\cos Q)^2 \\\\\\ \implies 1-2\sin Q+\sin^2 Q=16\cos^2 Q \\\\\\ \left[\textsf{Use }\cos^2\theta = 1-\sin^2\theta \right] \\ \\ \\ \implies 1-2\sin Q+\sin^2 Q=16(1-\sin^2Q)\\\\\\\implies 1-2\sin Q+\sin^2Q=16-16\sin^2Q$

$\displaystyle \implies 17\sin^2Q-2\sin Q-15=0 \\\\\\ \implies 17\sin^2 Q-17\sin Q+15\sin Q-15=0 \\\\\\ \implies 17\sin Q(\sin Q-1)+15(\sin Q-1)=0\\\\\\ \implies (\sin Q-1)(17\sin Q+15)=0 \\\\\\ \implies \sin Q-1=0 \quad \text{OR} \quad 17\sin Q+15=0 \\\\\\ \implies \sin Q=1 \quad \text{OR} \quad \sin Q= -\frac{15}{17}$

This is not a valid statement! So:-

$\displaystyle \sin Q = -\frac{15}{17} \\ \\ \\ \textsf{Use the identity: } \cos^2 \theta = 1-\sin^2\theta \\ \\ \\ \cos^2 Q=1-\left(-\frac{15}{17}\right)^2 \\\\\\ \implies \cos^2 Q = 1-\frac{225}{289} \\\\\\ \implies \cos^2 Q=\frac{289-225}{289}\\\\\\\implies \cos^2 Q=\frac{64}{289} \\\\\\ \implies \cos Q = \pm \frac{8}{17}$

$\displaystyle \begin{aligned}\sec Q-\tan Q &= -\frac{17}{8}-\left(\frac{-\frac{15}{17}}{-\frac{8}{17}}\right) \\ \\ \\ &= -\frac{17}{8}-\frac{15}{8} \\ \\ \\ &= \frac{-32}{8} \\ \\ \\ &= -4 \end{aligned}$

The answer must be positive!

So cos Q=8/17

This answer is original, not copied.

I have only borrowed some codes from @QGP sir.