If-sin-m-2-2mn-m-2-2mn-2n-2-then-prove-that-tan-m-2-2mn-2mn-2n-2-

Question Number 206970 by MATHEMATICSAM last updated on 02/May/24

If \sin θ = \frac{m^{2} + 2 m n}{m^{2} + 2 m n + 2 n^{2}} then prove

that \tan θ = \frac{m^{2} + 2 m n}{2 m n + 2 n^{2}} .

Answered by Rasheed.Sindhi last updated on 02/May/24

\sin θ = \frac{m^{2} + 2 m n}{m^{2} + 2 m n + 2 n^{2}}

\sin^{2} θ = {(\frac{m^{2} + 2 m n}{m^{2} + 2 m n + 2 n^{2}})}^{2}

1 - \sin^{2} θ = 1 - \frac{{(m^{2} + 2 m n)}^{2}}{{(m^{2} + 2 m n + 2 n^{2})}^{2}}

\cos^{2} θ = \frac{{(m^{2} + 2 m n + 2 n^{2})}^{2} - {(m^{2} + 2 m n)}^{2}}{{(m^{2} + 2 m n + 2 n^{2})}^{2}}

\tan^{2} θ = \frac{\sin^{2} θ}{\cos^{2}} = \frac{{(m^{2} + 2 m n)}^{2}}{{(m^{2} + 2 m n + 2 n^{2})}^{2}} \times \frac{{(m^{2} + 2 m n + 2 n^{2})}^{2}}{{(m^{2} + 2 m n + 2 n^{2})}^{2} - {(m^{2} + 2 m n)}^{2}}

\tan^{2} θ = \frac{{(m^{2} + 2 m n)}^{2}}{{(m^{2} + 2 m n + 2 n^{2}) - (m^{2} + 2 m n)} {(m^{2} + 2 m n + 2 n^{2}) + (m^{2} + 2 m n)}}

= \frac{{(m^{2} + 2 m n)}^{2}}{2 n^{2} (2 m^{2} + 4 m n + 2 n^{2})}

= \frac{{(m^{2} + 2 m n)}^{2}}{4 n^{2} (m^{2} + 2 m n + n^{2})}

= \frac{{(m^{2} + 2 m n)}^{2}}{4 n^{2} {(m + n)}^{2}}

\tan θ = \pm \frac{m (m + 2 n)}{2 n (m + n)}

Answered by Frix last updated on 02/May/24

t = \frac{s}{\sqrt{1 - s^{2}}}

\overset{}{s} = \frac{u}{v} \Rightarrow t = \frac{u}{\sqrt{v^{2} - u^{2}}}

u = m^{2} + 2 m n

v = m^{2} + 2 m n + 2 n^{2}

t = \frac{m (m + 2 n)}{2 ∣ (m + n) n ∣}

Facebook Tweet Pin