Solveeit Logo
Login

Question

Question: If \( \cot \theta + \cos ec\theta = 1.5 \) , then show that \( \cos \theta = \dfrac{5}{{13}} \) ....

If cot⁡θ+cos⁡ecθ=1.5\cot \theta + \cos ec\theta = 1.5 , then show that cos⁡θ=513\cos \theta = \dfrac{5}{{13}} .

Explanation

Solution

Hint : The given question involves solving a trigonometric equation and finding the value of angle θ\theta that satisfies the given equation. There can be various methods to solve a specific trigonometric equation. For solving such questions, we need to have knowledge of basic trigonometric formulae and identities.

Complete step-by-step answer :
The given problem requires us to solve the trigonometric equation cot⁡θ+cos⁡ecθ=1.5\cot \theta + \cos ec\theta = 1.5 .
The given trigonometric equation can be solved by first converting the trigonometric ratios into sine and cosine and then condensing them into trigonometric functions of compound angle.
So, we convert the given equation into sine and cosine,
⇒cos⁡θsin⁡θ+1sin⁡θ=32\Rightarrow \dfrac{{\cos \theta }}{{\sin \theta }} + \dfrac{1}{{\sin \theta }} = \dfrac{3}{2}
Simplifying the left side of the equation, we get,
⇒cos⁡θ+1sin⁡θ=32\Rightarrow \dfrac{{\cos \theta + 1}}{{\sin \theta }} = \dfrac{3}{2}
Cross multiplying the terms of the equation in order to simplify the trigonometric equation,
⇒2(cos⁡θ+1)=3sin⁡θ\Rightarrow 2\left( {\cos \theta + 1} \right) = 3\sin \theta
⇒2cos⁡θ+2=3sin⁡θ\Rightarrow 2\cos \theta + 2 = 3\sin \theta
Rearranging the terms,
⇒3sin⁡θ−2cos⁡θ=2\Rightarrow 3\sin \theta - 2\cos \theta = 2
Now, we know that sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1 . Hence, we get,
⇒±31−cos⁡2θ−2cos⁡θ=2\Rightarrow \pm 3\sqrt {1 - {{\cos }^2}\theta } - 2\cos \theta = 2
Rearranging the terms,
⇒±31−cos⁡2θ=2+2cos⁡θ\Rightarrow \pm 3\sqrt {1 - {{\cos }^2}\theta } = 2 + 2\cos \theta
Squaring both sides of the equation, we get,
⇒9(1−cos⁡2θ)=(2+2cos⁡θ)2\Rightarrow 9\left( {1 - {{\cos }^2}\theta } \right) = {\left( {2 + 2\cos \theta } \right)^2}
Opening the brackets and computing the square of binomial, we get,
⇒9−9cos⁡2θ=4+4cos⁡2θ+8cos⁡θ\Rightarrow 9 - 9{\cos ^2}\theta = 4 + 4{\cos ^2}\theta + 8\cos \theta
⇒13cos⁡2θ+8cos⁡θ−5=0\Rightarrow 13{\cos ^2}\theta + 8\cos \theta - 5 = 0
Now, solving the quadratic equation using the splitting the middle term method, we get,
⇒13cos⁡2θ+13cos⁡θ−5cos⁡θ−5=0\Rightarrow 13{\cos ^2}\theta + 13\cos \theta - 5\cos \theta - 5 = 0
⇒13cos⁡θ(cos⁡θ+1)−5(cos⁡θ+1)=0\Rightarrow 13\cos \theta \left( {\cos \theta + 1} \right) - 5\left( {\cos \theta + 1} \right) = 0
⇒(13cos⁡θ−5)(cos⁡θ+1)=0\Rightarrow \left( {13\cos \theta - 5} \right)\left( {\cos \theta + 1} \right) = 0
So, either (13cos⁡θ−5)=0\left( {13\cos \theta - 5} \right) = 0 or (cos⁡θ+1)=0\left( {\cos \theta + 1} \right) = 0,
Either cos⁡θ=513\cos \theta = \dfrac{5}{{13}} or cos⁡θ=−1\cos \theta = - 1
But if cos⁡θ=−1\cos \theta = - 1, then sin⁡θ=0\sin \theta = 0 which is not possible as cot⁡θ\cot \theta function would not be defined then.
So, the value of cos⁡θ\cos \theta is (513)\left( {\dfrac{5}{{13}}} \right) according to the condition or trigonometric equation given to us.
So, the correct answer is “ (513)\left( {\dfrac{5}{{13}}} \right) ”.

Note : Such trigonometric equations can be solved by various methods by applying suitable trigonometric identities and formulae. For solving such types of questions where we have to solve trigonometric equations, we need to have basic knowledge of algebraic rules and identities as well as a strong grip on trigonometric formulae and identities.