Question

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Answer 1

Answer:

$\dfrac{secx-tanx}{secx+tanx} = 1 -2secxtanx+2tan^{2}x$

Step-by-step explanation:

Given that-

$\dfrac{secx-tanx}{secx+tanx}$

By rationalization method-

= $\dfrac{secx-tanx}{secx+tanx}*\dfrac{secx-tanx}{secx-tanx}$

= $\dfrac{(secx-tanx)^{2} }{sec^{2}x-tan^{2}x  }$

= $\dfrac{sec^{2}x+tan^{2}x-2secxtanx }{sec^{2}x-tan^{2}x  }$

But $sec^{2}x = 1 + tan^{2}x$

And $sec^{2}x-tan^{2}x = 1$

= $\dfrac{(1+tan^{2}x)+tan^{2}x-2tanxsecx  }{1}$

= $1 + tan^{2}x+tan^{2}x-2secxtanx$

= $1 + 2tan^{2}x-2secxtanx$

Hence, $\dfrac{secx-tanx}{secx+tanx} = 1 -2secxtanx+2tan^{2}x$

Hence Proved