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Prove the following identity:
$fraction numerator cosecθ over denominator cosecθ minus 1 end fraction plus fraction numerator cosecθ over denominator cosecθ plus 1 end fraction equals 2 plus 2 tan squared straight theta.$

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392.

Prove the following identity:
$square root of fraction numerator 1 plus cosθ over denominator 1 minus cosθ end fraction end root equals space cosecθ plus cotθ.$

93 Views

Long Answer Type

393.

Prove the following identity:
$fraction numerator tanθ over denominator 1 minus cotθ end fraction plus fraction numerator cotθ over denominator 1 minus tanθ end fraction equals 1 plus tanθ plus cotθ.$

117 Views

Short Answer Type

394.

What is the value of (Cos²67^o - sin² 23^o)?

395.
If $4 \tan θ = 3, evaluate (\frac{4 \sin θ - \cos θ + 1}{4 \sin θ + \cos θ - 1})$

Given that,

$\tan θ = \frac{3}{4} ∴ \tan^{2} θ = \frac{9}{16} we know that, \sec^{2} θ = 1 + \tan^{2} θ ∴ \sec^{2} θ = 1 + \frac{9}{16} = \frac{25}{16} ∴ \cos^{2} θ = \frac{16}{25} ∴ \cos θ = \frac{4}{5} we know that, \sin^{2} θ = 1 - \cos^{2} θ ∴ \sin^{2} θ = 1 - \frac{16}{25} = \frac{9}{25} \Rightarrow \sin θ = \frac{3}{5} Now, (\frac{4 \sin θ - \cos θ + 1}{4 \sin θ + \cos θ - 1}) = (\frac{4 x (\frac{3}{5}) - \frac{4}{5} + 1}{4 x (\frac{3}{5}) + \frac{4}{5} - 1}) \frac{12 - 4 + 5}{12 + 4 - 5} = \frac{13}{11}$