We have find the double angle formula of sec(2x) in terms of cosec x and sec x
Solution
We know that \(\sec x = \frac{1}{cos x}\)
Hence, \(\sec 2x = \frac{1}{cos 2x}\) \(\sec 2x = \frac{1}{cos (x + x)}\)
We know the trigonometric identity cos(A+B)=cos(A)⋅cos(B)−sin(A)⋅sin(B)
On expanding{cos (x + x)} by using the above identity we get
= \(\frac{1}{\cos x . cos x + sin x .sin x}\)
= \(\frac{1}{\frac{1}{sec x}\frac{1}{sec x}+ \frac{1}{sin x}\frac{1}{sin x}}\)
= \(\frac{1}{\frac{\csc ^{2}x + \sec ^{2}x}{\sec ^{2}x.\sin ^{2}x}}\)
= \(\frac{sec^{2}x.csc^{2}x}{\csc ^{2}x + \sec ^{2}x}\)
Answer
sec(2x) formula in terms of cosec x and sec x is = \(\frac{sec^{2}x.csc^{2}x}{\csc ^{2}x + \sec ^{2}x}\)
