Prove that 1-tan 2 -4-A 1-tan 2 -4-A -sin bA Find b

1 tan ^2 ( π4 A )1+tan ^2 ( π4 A )=1 tan ^2θ1+tan ^2θ nbsp; nbsp;where π4 A=θ ) =cos 2θ =cos ( π/2 2A )=sin 2A Ans: 2 ...

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Relation between Inverses of Trigonometric Functions and Their Reciprocal Functions

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Prove that $\frac{1 - {tan}^{2} (\frac{π}{4} - A)}{1 + {tan}^{2} (\frac{π}{4} - A)} = sin b A$ Find b

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\frac{1 - t a n^{2} (\frac{π}{4} + θ)}{1 + t a n^{2} (\frac{π}{4} + θ)} =

{sin}^{2} \frac{π}{6} + {cos}^{2} \frac{π}{3} - {tan}^{2} \frac{π}{4} = - \frac{1}{2}

{sin}^{2} \frac{π}{6} + {cos}^{2} \frac{π}{3} - {tan}^{2} \frac{π}{4} = - \frac{1}{2}

\frac{1 - {tan}^{2} (\frac{π}{4} - A)}{1 + {tan}^{2} (\frac{π}{4} - A)}

A \in [0, \frac{π}{2}]

tan (\frac{π}{4} + x) + tan (\frac{π}{4} - x) = a

{tan}^{2} (\frac{π}{4} + x) + {tan}^{2} (\frac{π}{4} - x) =

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