\sec \theta - \tan \theta = x ⇒ \frac{1+x^2}{1-x^2} =

TS ICET 24 May 2019 Shift 1 Paper

Question : 119 of 200

Marks: +1, -0

sec θ - tan θ = x

⇒

{1+x^2}/{1-x^2}

Solution:

\sec θ-\tan θ=x

\;{1+x^{2}}/{1-x^{2}}=\;{1+\sec ^{2} θ+\tan ^{2} θ-2 \sec θ \tan θ}/{1-\sec ^{2} θ-\tan ^{2} θ+2 \sec θ \tan θ}

\;{2 \sec ^{2} θ-2 \sec θ \tan θ}/{-2 \tan ^{2} θ+2 \sec θ \tan θ}

⇒ \;{2 \sec θ(\sec θ-\tan θ)}/{2 \tan θ(\sec θ-\tan θ)}

=\;{1}/{\sin θ}=\cosec θ

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