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

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

Explanation

Solution

To find the value of cos⁡θ\cos \theta we will find the roots of the given equation. To find the roots of the given equation, we will rewrite tan⁡2θ{\tan ^2}\theta as sec⁡2θ−1{\sec ^2}\theta - 1 . Then we can write sec⁡θ\sec \theta as xx to write the expression as a quadratic equation. So, we can find the roots of the quadratic equation by either factorization method or by the quadratic formula. We know that sec⁡θ\sec \theta is the inverse of cos⁡θ\cos \theta . Hence, we can find the value of cos⁡θ\cos \theta .

Complete step by step solution:
Given:
The given expression is tan⁡2θ+sec⁡θ=5{\tan ^2}\theta + \sec \theta = 5.

We know from the basics of trigonometry that we can write sec⁡2θ−1{\sec ^2}\theta - 1 for tan⁡2θ{\tan ^2}\theta . We can rewrite the given expression as:
sec⁡2θ−1+sec⁡θ=5 sec⁡2θ+sec⁡θ−1−5=0 sec⁡2θ+sec⁡θ−6=0\begin{array}{c} {\sec ^2}\theta - 1 + \sec \theta = 5\\\ {\sec ^2}\theta + \sec \theta - 1 - 5 = 0\\\ {\sec ^2}\theta + \sec \theta - 6 = 0 \end{array}

Now, we have to solve the above equation, we will write xx for sec⁡θ\sec \theta .
x2+x−6=0{x^2} + x - 6 = 0

The above equation is the quadratic equation. We will find the roots of the quadratic equation by factorization method.
x2+3x−2x−6=0 x(x+3)−2(x+3)=0 (x−2)(x+3)=0\begin{array}{c} {x^2} + 3x - 2x - 6 = 0\\\ x\left( {x + 3} \right) - 2\left( {x + 3} \right) = 0\\\ \left( {x - 2} \right)\left( {x + 3} \right) = 0 \end{array}

From the above expression, we have two roots of the quadratic equation which can be expressed as:
x−2=0 x=2\begin{array}{c} x - 2 = 0\\\ x = 2 \end{array}

And another root of quadratic equation is,
x+3=0 x=−3\begin{array}{c} x + 3 = 0\\\ x = - 3 \end{array}

Now, we will again substitute sec⁡θ\sec \theta for xx we get,
sec⁡θ=2\sec \theta = 2 and sec⁡θ=−3\sec \theta = - 3

We know that sec⁡θ\sec \theta is the inverse of the cos⁡θ\cos \theta . Therefore, we can rewrite the roots as:

1cos⁡θ=2   cos⁡θ=12\begin{array}{l} \dfrac{1}{{\cos \theta }} = 2\;\\\ \cos \theta = \dfrac{1}{2} \end{array}
And,
1cos⁡θ=−3     cos⁡θ=−13\begin{array}{l} \dfrac{1}{{\cos \theta }} = - 3\;\\\ \;\cos \theta = - \dfrac{1}{3} \end{array}
Hence we have cos⁡θ=12  or  −13\cos \theta = \dfrac{1}{2}\;{\rm{or}}\; - \dfrac{1}{3}.

Note: We should have prior knowledge about the basic trigonometric formulas. The conversion of the trigonometric function from one to another is the key point to solve this problem. Hence, we need to carefully substitute the values in the expression. Also, the equation formed is a quadratic equation whose roots we need to find. Here we are using the factorization method but we can also use a quadratic formula to get our solution.