"Verify the following identity:tan⁡x1+sec⁡x+1+sec⁡xtan⁡x=2csc⁡x" solved and explained

Rupert Albert

Answered question

2022-03-01

Verify the following identity:

\frac{\tan x}{1 + \sec x} + \frac{1 + \sec x}{\tan x} = 2 \csc x

Rosalind Barker

Beginner2022-03-02Added 7 answers

Method 1:
As

\tan^{2} x = \sec^{x} - 1 = (\sec x - 1) (\sec x + 1), \frac{\tan x}{\sec x - 1}

= \frac{\sec x + 1}{\tan x}

\Rightarrow \frac{\tan x}{1 + \sec x} + \frac{1 + \sec x}{\tan x} = \frac{\sec x - 1}{\tan x} + \frac{1 + \sec x}{\tan x}

= \frac{2 \sec x}{\tan x} = \frac{\frac{2}{\cos x}}{\frac{\sin x}{\cos x}} = \frac{2}{\sin x}

Method 2:

\frac{\tan x}{1 + \sec x} = \frac{\frac{\sin x}{\cos x}}{1 + \frac{1}{\cos x}} = \frac{\sin x}{1 + \cos x} = \frac{\sin x (1 - \cos x)}{(1 + \cos x) (1 - \cos x)}

= \frac{1 - \cos x}{\sin x}

\sin^{2} x = 1 - \cos^{2} x

and

\frac{1 + \sec x}{\tan x} = \frac{1 + \frac{1}{\cos x}}{\frac{\sin x}{\cos x}} = \frac{1 + \cos x}{\sin x}

Nathan Kent

Beginner2022-03-03Added 8 answers

\frac{\tan x}{1 + \sec x} + \frac{1 + \sec x}{\tan x}

= \frac{\tan^{2} x + {(1 + \sec x)}^{2}}{(1 + \sec x) \tan x}

= \frac{\tan^{2} x + 1 + \sec^{2} x + 2 \sec x}{(1 + \sec x) \tan x}

= \frac{\sec^{2} x + \sec^{2} x + 2 \sec x}{(1 + \sec x) \tan x}

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

= \frac{2 \sec x}{\tan x}

= \frac{\frac{2}{\cos x}}{\frac{\sin x}{\cos x}}

= \frac{2}{\cos x} \times \frac{\cos x}{\sin x}

= \frac{2}{\sin x}

= 2 \csc x

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