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ExtremeBDS [4]
3 years ago
12

Prove the identity

Mathematics
1 answer:
Troyanec [42]3 years ago
3 0

Answer:

Proved

Step-by-step explanation:

The options are not given. So, I will solve from scratch

Given

\frac{secx-1}{tan x}= \frac{tanx}{secx+1}

Required

Prove

Multiply the right-hand side by \frac{secx + 1}{secx + 1}

\frac{secx-1}{tan x} * \frac{secx + 1}{secx + 1}= \frac{tanx}{secx+1}

Apply difference of two squares on the numerator

\frac{sec^2 x - 1}{(tanx)(secx + 1)} =\frac{tanx}{secx+1}

In trigonometry:

tan^2x = sec^2x - 1

So, we have:

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

\frac{tan x * tan x}{(tanx)(secx + 1)} =\frac{tanx}{secx+1}

tan x cancels out

\frac{tan x}{secx + 1} =\frac{tanx}{secx+1}

<em>Proved</em>

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