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Complex Numbers

If tanθ+θ=m...

Question

If $tan θ + sec θ = m$ and $tan θ - sec θ = n$ Show that $m^{2} - n^{2} = 2 tan θ . s e c θ$

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Solution

We have
$\begin{matrix} tan θ + sec θ = m - - - - (1) tan θ - sec θ = n - - - - - (2) a d d i n g (1) a n d (2) w e g e t (m + n) = 2 tan θ - - - - (3) s u b t r a c t i n g (2) f r o m (1) w e g e t (m - n) = 2 sec θ - - - - (4) m u l t i p l y i n g (3) a n d (4) (m + n) (m - n) = 2 tan θ . sec θ ∴ m^{2} - n^{2} = 2 tan θ . s e c θ p r o v e d . \end{matrix}$

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Similar questions

m sin θ = n cos θ

then show that :

\frac{tan θ + cot θ}{tan θ - cot θ} = \frac{n sin θ + m cos θ}{n sin θ - m cos θ} = \frac{n^{2} + m^{2}}{n^{2} - m^{2}}

Q. If (sec θ + tan θ) = m and (sec θ − tan θ) = n, show that mn = 1.

\tan θ + \sin θ = m a n d \tan θ - \sin θ = n, s h o w t h a t m^{2} + n^{2} = 4 \sqrt{m n} .

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

, determine the values of

tan θ

sec θ

csc θ

tan θ = \frac{m}{n},

θ = \frac{K}{L}

, then find the value of

\frac{L^{2} - K^{2}}{L^{2} n^{2}} \times (m^{2} + n^{2})

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