How do you find a double angle formula for sec (2x) in terms of only cosec(x) and sec(x)?

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 = \(\frac{sec^{2}x.csc^{2}x}{\csc ^{2}x + \sec ^{2}x}\)

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