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Question: If \({\tan ^2}\theta + \sec \theta = 5\) , find \(\cos \theta \) ....

If tan⁡2θ+sec⁡θ=5{\tan ^2}\theta + \sec \theta = 5 , find cos⁡θ\cos \theta .

Explanation

Solution

Firstly, use tan⁡2θ=sec⁡2θ−1{\tan ^2}\theta = {\sec ^2}\theta - 1 . After that, a quadratic equation will be formed. By solving that quadratic equation, using the method of splitting the middle term, we will get max. two values of sec⁡θ\sec \theta . And by taking the reciprocal of sec⁡θ\sec \theta , we will get the value of cos⁡θ\cos \theta .

Complete step-by-step answer:
We have
tan⁡2θ+sec⁡θ=5{\tan ^2}\theta + \sec \theta = 5
We know that, tan⁡2θ=sec⁡2θ−1{\tan ^2}\theta = {\sec ^2}\theta - 1
∴sec⁡2θ−1+sec⁡θ−5=0 ∴sec⁡2θ+sec⁡θ−6=0  \therefore {\sec ^2}\theta - 1 + \sec \theta - 5 = 0 \\\ \therefore {\sec ^2}\theta + \sec \theta - 6 = 0 \\\
Now, on splitting the middle term sec⁡θ\sec \theta as 3sec⁡θ−2sec⁡θ3\sec \theta - 2\sec \theta we get,
sec⁡2θ+3sec⁡θ−2sec⁡θ−6=0 ∴sec⁡θ(sec⁡θ+3)−2(sec⁡θ+3)=0 ∴(sec⁡θ+3)(sec⁡θ−2)=0  {\sec ^2}\theta + 3\sec \theta - 2\sec \theta - 6 = 0 \\\ \therefore \sec \theta \left( {\sec \theta + 3} \right) - 2\left( {\sec \theta + 3} \right) = 0 \\\ \therefore \left( {\sec \theta + 3} \right)\left( {\sec \theta - 2} \right) = 0 \\\
∴sec⁡θ+3=0\therefore \sec \theta + 3 = 0 or sec⁡θ−2=0\sec \theta - 2 = 0
∴sec⁡θ=−3\therefore \sec \theta = - 3 or sec⁡θ=2\sec \theta = 2
∴1cos⁡θ=−3\therefore \dfrac{1}{{\cos \theta }} = - 3 or 1cos⁡θ=2\dfrac{1}{{\cos \theta }} = 2
∴cos⁡θ=−13\therefore \cos \theta = \dfrac{{ - 1}}{3} or cos⁡θ=12\cos \theta = \dfrac{1}{2}
Thus, cos⁡θ=−13\cos \theta = \dfrac{{ - 1}}{3} or cos⁡θ=12\cos \theta = \dfrac{1}{2} .

Note: Alternate Method:
We have
tan⁡2θ+sec⁡θ=5{\tan ^2}\theta + \sec \theta = 5
We know that, tan⁡2θ=sec⁡2θ−1{\tan ^2}\theta = {\sec ^2}\theta - 1
∴sec⁡2θ−1+sec⁡θ−5=0 ∴sec⁡2θ+sec⁡θ−6=0  \therefore {\sec ^2}\theta - 1 + \sec \theta - 5 = 0 \\\ \therefore {\sec ^2}\theta + \sec \theta - 6 = 0 \\\
Now, from the above equation, a=1a = 1 , b=1b = 1 and c=−6c = - 6 .
∴Δ=b2−4ac\therefore \Delta = \sqrt {{b^2} - 4ac}
=12−4(1)(−6) =1+24 =25 =5  = \sqrt {{1^2} - 4\left( 1 \right)\left( { - 6} \right)} \\\ = \sqrt {1 + 24} \\\ = \sqrt {25} \\\ = 5 \\\
So, the roots of the equation are sec⁡θ=−b±Δ2a\sec \theta = \dfrac{{ - b \pm \Delta }}{{2a}} .
∴sec⁡θ=−b+Δ2a\therefore \sec \theta = \dfrac{{ - b + \Delta }}{{2a}} or sec⁡θ=−b−Δ2a\sec \theta = \dfrac{{ - b - \Delta }}{{2a}}
∴sec⁡θ=−1+52\therefore \sec \theta = \dfrac{{ - 1 + 5}}{2} or sec⁡θ=−1−52\sec \theta = \dfrac{{ - 1 - 5}}{2}
∴sec⁡θ=42\therefore \sec \theta = \dfrac{4}{2} or sec⁡θ=−62\sec \theta = \dfrac{{ - 6}}{2}
∴sec⁡θ=2\therefore \sec \theta = 2 or sec⁡θ=−3\sec \theta = - 3
∴1cos⁡θ=2\therefore \dfrac{1}{{\cos \theta }} = 2 or 1cos⁡θ=−3\dfrac{1}{{\cos \theta }} = - 3
∴cos⁡θ=12\therefore \cos \theta = \dfrac{1}{2} or cos⁡θ=−13\cos \theta = \dfrac{{ - 1}}{3}
Thus, cos⁡θ=−13\cos \theta = \dfrac{{ - 1}}{3} or cos⁡θ=12\cos \theta = \dfrac{1}{2} .