Key to "How to find a double angle formula for..." - Plainmath

mpifanahhf

Answered question

2023-02-15

How to find a double angle formula for

\sec 2 x

in terms of only

cosec x

and

\sec x

?

Answer & Explanation

abuhishcy

Beginner2023-02-16Added 6 answers

\sec 2 x = \frac{1}{\cos 2 x}

[∵ secx = \frac{1}{cosx}]

\Rightarrow \sec 2 x = \frac{1}{\cos (x + x)}

\Rightarrow \sec 2 x = \frac{1}{\cos x \cdot \cos x - \sin x \cdot \sin x}

[∵ \cos (A + B) = cosA \cdot cosB - sinA \cdot sinB]

\Rightarrow \sec 2 x = \frac{1}{\frac{1}{\sec x \cdot \sec x} - \frac{1}{cosec x \cdot c osec x}}

[∵ cosx = \frac{1}{secx}, sinx = \frac{1}{cosecx}]

\Rightarrow \sec 2 x = \frac{1}{\frac{1}{\sec^{2} x} - \frac{1}{{cosec}^{2} x}}

\Rightarrow \sec 2 x = \frac{1}{\frac{{cosec}^{2} x - \sec^{2} x}{\sec^{2} x \cdot {cosec}^{2} x}}

\Rightarrow \sec 2 x = \frac{\sec^{2} x \cdot {cosec}^{2} x}{{cosec}^{2}}

[∵ \frac{1}{\frac{a}{b}} = \frac{b}{a}]

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