Question

Prove that: $\cos ec\left( {\dfrac{\pi }{4} + \theta } \right)\cos ec\left( {\dfrac{\pi }{4} - \theta } \right) = \sec \left( {\dfrac{\pi }{4} + \theta } \right)\sec \left( {\dfrac{\pi }{4} - \theta } \right) = 2\sec 2\theta $

Answer 1

Hint: Before attempting this question prior knowledge of trigonometric identities is must, remember to use trigonometric identities $\sec \theta = \dfrac{1}{{\cos \theta }}$ use this information to approach the solution of the question.

Complete step-by-step solution:
According to the given information we have trigonometric equation $\cos ec\left( {\dfrac{\pi }{4} + \theta } \right)\cos ec\left( {\dfrac{\pi }{4} - \theta } \right) = \sec \left( {\dfrac{\pi }{4} + \theta } \right)\sec \left( {\dfrac{\pi }{4} - \theta } \right) = 2\sec 2\theta $ to prove the L.H.S = R.H.S
Let first prove $\sec \left( {\dfrac{\pi }{4} + \theta } \right)\sec \left( {\dfrac{\pi }{4} - \theta } \right) = 2\sec 2\theta $ here L.H.S is $\sec \left( {\dfrac{\pi }{4} + \theta } \right)\sec \left( {\dfrac{\pi }{4} - \theta } \right)$ and R.H.S is $2\sec 2\theta $
Since we know that $\sec \theta = \dfrac{1}{{\cos \theta }}$
Therefore $\sec \left( {\dfrac{\pi }{4} + \theta } \right)\sec \left( {\dfrac{\pi }{4} - \theta } \right) = \dfrac{1}{{\cos \left( {\dfrac{\pi }{4} + \theta } \right)\cos \left( {\dfrac{\pi }{4} - \theta } \right)}}$
L.H.S = $\dfrac{1}{{\cos \left( {\dfrac{\pi }{4} + \theta } \right)\cos \left( {\dfrac{\pi }{4} - \theta } \right)}}$
Multiplying and dividing by 2 we get
L.H.S = $\dfrac{1}{{\cos \left( {\dfrac{\pi }{4} + \theta } \right)\cos \left( {\dfrac{\pi }{4} - \theta } \right)}} \times \dfrac{2}{2}$
L.H.S = $\dfrac{1}{{\cos \left( {\dfrac{\pi }{4} + \theta } \right)\cos \left( {\dfrac{\pi }{4} - \theta } \right)}} \times \dfrac{2}{2}$
By the formula 2 Cos A Cos B = Cos (A+B) + Cos (A – B) we get
 L.H.S = $\dfrac{2}{{\cos \left( {\dfrac{\pi }{4} + \theta + \dfrac{\pi }{4} - \theta } \right) + \cos \left( {\dfrac{\pi }{4} + \theta - \dfrac{\pi }{4} + \theta } \right)}}$
L.H.S = $\dfrac{2}{{\cos \dfrac{\pi }{2} + \cos 2\theta }}$
Since we know that $\cos \dfrac{\pi }{2}$ = 0
Therefore L.H.S = $\dfrac{2}{{\cos 2\theta }}$
Since we know that $\sec \theta = \dfrac{1}{{\cos \theta }}$
So, L.H.S = $2\sec 2\theta $ (equation 1)
So, L.H.S is equal to R.H.S
Now taking $\cos ec\left( {\dfrac{\pi }{4} + \theta } \right)\cos ec\left( {\dfrac{\pi }{4} - \theta } \right)$ as L.H.S and $2\sec 2\theta $ as R.H.S
Since we know that $\cos ec\theta = \dfrac{1}{{\sin \theta }}$
Therefore L.H.S = $\cos ec\left( {\dfrac{\pi }{4} + \theta } \right)\cos ec\left( {\dfrac{\pi }{4} - \theta } \right) = \dfrac{1}{{\sin \left( {\dfrac{\pi }{4} + \theta } \right)\sin \left( {\dfrac{\pi }{4} - \theta } \right)}}$
Multiplying and dividing by 2 we get
L.H.S = $\dfrac{1}{{\sin \left( {\dfrac{\pi }{4} + \theta } \right)\sin \left( {\dfrac{\pi }{4} - \theta } \right)}} \times \dfrac{2}{2}$
Since we know that 2 Sin A Sin B = Cos (A – B) – Cos (A + B)
Therefore L.H.S = $\dfrac{2}{{\cos \left( {\dfrac{\pi }{4} + \theta - \dfrac{\pi }{4} + \theta } \right) - \cos \left( {\dfrac{\pi }{4} + \theta + \dfrac{\pi }{4} - \theta } \right)}}$
$ \Rightarrow $L.H.S = $\dfrac{2}{{\cos 2\theta - \cos \dfrac{\pi }{2} }}$
Since we know that $\cos \dfrac{\pi }{2}$ = 0
Therefore L.H.S = $\dfrac{2}{{\cos 2\theta }}$
Since we know that $\sec \theta = \dfrac{1}{{\cos \theta }}$
So, L.H.S = $2\sec 2\theta $ (equation 2)
So, L.H.S is equal to R.H.S
By the equation 1 and equation 2 we can say that
$\cos ec\left( {\dfrac{\pi }{4} + \theta } \right)\cos ec\left( {\dfrac{\pi }{4} - \theta } \right) = \sec \left( {\dfrac{\pi }{4} + \theta } \right)\sec \left( {\dfrac{\pi }{4} - \theta } \right)$
Hence $\cos ec\left( {\dfrac{\pi }{4} + \theta } \right)\cos ec\left( {\dfrac{\pi }{4} - \theta } \right) = \sec \left( {\dfrac{\pi }{4} + \theta } \right)\sec \left( {\dfrac{\pi }{4} - \theta } \right) = 2\sec 2\theta $ is proved

Note: The above question was totally based on the concept of trigonometry and its identities which can be explained as the concept which relates the sides and the angle of a right-angled triangle whereas the trigonometric identities are equalities which consist of trigonometric identities like sin theta, cos theta, etc. which are true for every occurring variable in such cases where sides of both are well defined.