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Question Number 170336 by Mastermind last updated on 21/May/22

i f {c o s}^{2} \propto + 3 {c o s e c}^{2} θ = 7, f i n d t h e

e x p r e s s i o n o f {t a n}^{2} θ, H e n c e s h o w t h a t

t a n θ = 1 .

M a s t e r m i n d

Commented by MATHSLORD22 last updated on 21/May/22

\cos^{2} θ or \cos^{2} α ? ?

Answered by thfchristopher last updated on 21/May/22

\cos^{2} θ + {3 \csc}^{2} θ = 7

\Rightarrow \sin^{2} θ \cos^{2} θ + 3 = {7 \sin}^{2} θ

\Rightarrow \sin^{2} θ (1 - \sin^{2} θ) + 3 = {7 \sin}^{2} θ

\Rightarrow \sin^{2} θ - \sin^{4} θ + 3 = {7 \sin}^{2} θ

\Rightarrow \sin^{4} θ + {6 \sin}^{2} θ - 3 = 0

\sin^{2} θ = \frac{- 6 \pm \sqrt{36 + 12}}{2} = \frac{- 6 \pm 4 \sqrt{3}}{2}

∴ \sin^{2} θ = 2 \sqrt{3} - 3

\Rightarrow \cos^{2} θ = 4 - 2 \sqrt{3}

\Rightarrow \tan^{2} θ = \frac{2 \sqrt{3} - 3}{4 - 2 \sqrt{3}}

= \frac{\sqrt{3}}{2}

Commented by MATHSLORD22 last updated on 21/May/22

He wrote α not θ

Commented by mr W last updated on 21/May/22

He wrote \propto not α :)

Commented by Mastermind last updated on 21/May/22

T h a n k y o u m a n

Answered by 2kdw last updated on 28/Apr/23

i) {c o s}^{2} α + \frac{3}{{s i n}^{2} α} = 7

i f {c o s}^{2} x = 1 - {s i n}^{2} x \dots (1)

a n d {s i n}^{2} x = t \dots (2)

∴ t (1 - t) + 3 = 7 t

- t^{2} + t + 3 = 7 t

- t^{2} - 6 t + 3 = 0

∴ {(t + 3)}^{2} = 12

\Rightarrow t = 2 \sqrt{3} - 3 o r t = - 3 - 2 \sqrt{3}

b u t (1) 1 - t ⩾ 0 t h e n t ⩽ 1

∴ \begin{matrix} t = 2 \sqrt{3} - 3 \end{matrix}

i i) I f (2) t = {s i n}^{2} α

[{s i n}^{2} α = 2 \sqrt{3} - 3]

\dots (1) {c o s}^{2} α = 1 - 2 \sqrt{3} + 3

∴ {t a n}^{2} α = \frac{2 \sqrt{3} - 3}{1 - 2 \sqrt{3} + 3}

\begin{matrix} {t a n}^{2} α = \frac{2 \sqrt{3} - 3}{4 - 2 \sqrt{3}} \end{matrix}

Commented by Mastermind last updated on 21/May/22

T h a n k s m a n