m-2-sin-2m-1-cos-2m-1-if-1-tan-3-4-2-tan-4-3-3-tan-2m-m-2-1-4-tan-2m-m-2-1-

Question Number 18184 by Tinkutara last updated on 16/Jul/17

(m + 2) \sin θ + (2 m - 1) \cos θ = 2 m + 1, if

(1) \tan θ = \frac{3}{4}

(2) \tan θ = \frac{4}{3}

(3) \tan θ = \frac{2 m}{m^{2} - 1}

(4) \tan θ = \frac{2 m}{m^{2} + 1}

Commented by prakash jain last updated on 16/Jul/17

\tan \frac{θ}{2} = u

\sin θ = \frac{2 u}{1 + u^{2}}, \cos θ = \frac{1 - u^{2}}{1 + u^{2}}

(m + 2) 2 u + (2 m - 1) (1 - u^{2}) = (2 m + 1) (1 + u^{2})

4 {m u}^{2} + 2 (m + 2) u + 2 = 0

2 {m u}^{2} - (m + 2) u + 1 = 0

(2 u - 1) (m u - 1) = 0

u = \frac{1}{2}, \frac{1}{m}

\tan θ = \frac{2 \tan θ / 2}{1 - \tan^{2} θ / 2}

\tan θ = \frac{4}{3}, \frac{2 m}{m^{2} - 1}

Commented by Tinkutara last updated on 17/Jul/17

Thanks Sir!

Answered by ajfour last updated on 16/Jul/17

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

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

= - \tan^{- 1} \frac{2 m - 1}{m + 2} + \tan^{- 1} \frac{2 m + 1}{m - 2}

\tan θ = \frac{\frac{2 m + 1}{m - 2} - \frac{2 m - 1}{m + 2}}{1 + \frac{2 m + 1}{m - 2} . \frac{2 m - 1}{m + 2}}

= \frac{(2 m + 1) (m + 2) - (2 m - 1) (m - 2)}{m^{2} - 4 + {4 m}^{2} - 1}

\tan θ = \frac{10 m}{5 (m^{2} - 1)} = \frac{2 m}{m^{2} - 1} .

[option (3), (2)]

for m = 2, \tan θ = \frac{4}{3} .

Commented by ajfour last updated on 16/Jul/17

you can view solution of Q . 18111

(chord AB at ∠ = ϕ with diameter

cut in the ratio p : q)

Commented by Tinkutara last updated on 16/Jul/17

I viewed it . Nice method! And thanks

for this solution!