If-asin-bcos-2ctan-1-tan-2-then-prove-that-a-2-b-2-2-4c-2-a-2-b-2-

Question Number 206471 by MATHEMATICSAM last updated on 15/Apr/24

If a \sin θ = b \cos θ = \frac{2 c \tan θ}{1 - \tan^{2} θ} then prove

that {(a^{2} - b^{2})}^{2} = 4 c^{2} (a^{2} + b^{2}) .

Answered by lepuissantcedricjunior last updated on 15/Apr/24

a s i n θ = \frac{2 c t a n θ}{1 - {t a n}^{2} θ} = c t a n 2 θ = \frac{2 c s i n θ c o s θ}{2 {c o s}^{2} θ - 1} = b c o s θ

\Leftrightarrow a = \frac{2 c c o s θ}{2 {c o s}^{2} θ - 1}; b = \frac{2 c s i n θ}{2 {c o s}^{2} θ - 1}

o n a

{(a^{2} - b^{2})}^{2} = {(\frac{4 c^{2} c o s 2 θ}{{(2 {c o s}^{2} θ - 1)}^{2}})}^{2} = 16 c^{4} {(\frac{c o s 2 θ}{{c o s}^{2} (2 θ)})}^{2}

\Rightarrow {(a^{2} - b^{2})}^{2} = 16 c^{4} (\frac{1}{{c o s}^{2} (2 θ)})

= 16 c^{4} [\frac{1}{4 c^{2}} (\frac{4 c^{2} {s i n}^{2} θ}{{c o s}^{2} (2 θ)} + \frac{4 c^{2} {c o s}^{2} θ}{{c o s}^{2} (2 θ)})]

= 4 c^{2} (a^{2} + b^{2})

\Rightarrow {(a^{2} - b^{2})}^{2} = 4 c^{2} (a^{2} + b^{2})

\dots \dots \dots l e p u i s s a n t c e d r i c j u n i o r \dots . .