if-q-1-sin-then-prove-that-sec-tan-2-1-q-

Question Number 159845 by abdullah_ff last updated on 21/Nov/21

if q = 1 - s i n θ; then prove that

{(s e c θ - t a n θ)}^{2} = \frac{1}{q}

Commented by tounghoungko last updated on 21/Nov/21

\sin θ = 1 - q

\cos θ = \sqrt{1 - (1 - 2 q + q^{2})} = \sqrt{2 q - q^{2}}

\sec θ = \frac{1}{\sqrt{2 q - q^{2}}}; \tan θ = \frac{1 - q}{\sqrt{2 q - q^{2}}}

\sec θ - \tan θ = \frac{q}{\sqrt{2 q - q^{2}}}

{(\sec θ - \tan θ)}^{2} = \frac{q^{2}}{2 q - q^{2}}

Answered by JDamian last updated on 21/Nov/21

{(\frac{1 - s i n θ}{c o s θ})}^{2} = \frac{q^{2}}{{c o s}^{2} θ} = \frac{q^{2}}{1 - {s i n}^{2} θ} = \frac{q^{2}}{1 - {(1 - q)}^{2}} =

\frac{q^{2}}{1 - 1 - q^{2} + 2 q} = \frac{q}{2 - q}

Answered by som(math1967) last updated on 21/Nov/21

{(s e c θ - t a n θ)}^{2}

= {(\frac{1 - s i n θ}{c o s θ})}^{2}

= \frac{{(1 - s i n θ)}^{2}}{{c o s}^{2} θ}

= \frac{{(1 - s i n θ)}^{2}}{(1 + s i n θ) (1 - s i n θ)}

= \frac{1 - s i n θ}{1 + s i n θ} \neq \frac{1}{q}