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Question: If \(\cos ec\theta + \cot \theta = m\) and \(\cos ec\theta - \cot \theta = n\) , prove that mn=1....

If cos⁡ecθ+cot⁡θ=m\cos ec\theta + \cot \theta = m and cos⁡ecθ−cot⁡θ=n\cos ec\theta - \cot \theta = n , prove that mn=1.

Explanation

Solution

Hint: Here we have to simplify the equations using trigonometry identities to find the product of both equations.

Complete step-by-step answer:
Given cos⁡ecθ+cot⁡θ=m\cos ec\theta + \cot \theta = m and cos⁡ecθ−cot⁡θ=n\cos ec\theta - \cot \theta = n
To prove: mn=1
Taking LHS, mn=(cos⁡ecθ+cot⁡θ)(cos⁡ecθ−cot⁡θ)mn = \left( {\cos ec\theta + \cot \theta } \right)\left( {\cos ec\theta - \cot \theta } \right)
Using formula (a−b)(a+b)=a2−b2\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}
⇒\Rightarrow mn=cos⁡ec2θ−cot⁡2θmn = \cos e{c^2}\theta - {\cot ^2}\theta
We know that cos⁡ec2θ=1+cot⁡2θ⇒cos⁡ec2θ−cot⁡2θ=1\cos e{c^2}\theta = 1 + {\cot ^2}\theta \Rightarrow \cos e{c^2}\theta - {\cot ^2}\theta = 1
⇒\Rightarrow mn=1=RHS
Hence Proved.

Note: These types of problems can be easily solved with the help of understanding of trigonometric identities and formulas.