Question

prove that secФ-tanФ/secФ+tanФ=1-2sec.tanФ+2tan²Ф.

Answer 1

$X=\frac{sec\theta-tan\theta}{sec\theta+tan\theta}$

   Let us multiply the denominator and numerator with a rationalizing factor.  So that we can simplify the denominator. 

We know that sec² θ - tan² θ = 1 
             or, (sec θ - tanθ) (sec θ + tan θ) = 1,    as  A² - B² = (A-B) (A+B)

$X=\frac{(sec\theta-tan\theta)^2}{(sec\theta+tan\theta)(sec\theta-tan\theta)}\\\\=\frac{sec^2\theta+tan^2\theta-2sec\theta\ tan\theta}{sec^2\theta-tan^2\theta}\\\\=1+tan^2\theta+tan^2\theta-2sec\theta\tan\theta=\\\\=1-2\ sec\theta\ tan\theta+2\ tan^2\theta$

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similarly we need to use:  Cosec² Ф - Cot² Ф = 1
   Sin² Ф + Cos ²Ф = 1    while rationalizing  fractions.
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if you are comfortable with Sin Ф and Cos Ф, then:

$\frac{sec\theta-tan\theta}{sec\theta+tan\theta}\\\\=\frac{\frac{1}{cos\theta}-\frac{sin\theta}{cos\theta}}{\frac{1}{cos\theta}+\frac{sin\theta}{cos\theta}}\\\\=\frac{1-sin\theta}{1+sin\theta}\\\\multiply\ with\ (1-sin\theta)\ numerator\ and\ denominator\\\\=\frac{(1-sin\theta)^2}{1^2-sin^2\theta}\\\\=\frac{1+sin^2\theta-2sin\theta}{cos^2\theta}\\\\=\frac{1}{cos^2\theta}+\frac{sin^2\theta}{cos^2\theta}-\frac{2\sin\theta}{cos\theta\ cos\theta}\\\\=sec^2\theta+tan^2\theta-2tan\theta\ sec\theta\\\\.....$