If sin - - a2 - b2-a2 - b2- find the values of all T-ratios of -

Sol: In ΔABC, Let ∠ABC be θ sin θ = a^2 b^2/a^2 + b^2 ⇒ AB = (a^2 b^2) ⇒ AC = (a^2 + b^2) ⇒ BC = √((a^2 + b^2)^2 (a^2 b^2)^2) [According to Pythagora ...

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Standard X

Mathematics

Relation between Trigonometric Ratios

If sin θ = a2...

Question

If sin $θ = \frac{a^{2} - b^{2}}{a^{2} + b^{2}}$ , find the values of all T-ratios of $θ$

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Solution

Sol:

In ΔABC, Let $∠$ ABC be $θ$

sin $θ = \frac{a^{2} - b^{2}}{a^{2} + b^{2}}$

⇒ AB = $(a^{2} - b^{2})$

⇒ AC = $(a^{2} + b^{2})$

⇒ BC = $\sqrt{(a^{2} + b^{2})^{2} - (a^{2} - b^{2})^{2}}$ [According to Pythagoras theorem]

⇒ BC = $\sqrt{4 a^{2} b^{2}}$

⇒ BC = $2 a b$

$c o s θ = \frac{2 a b}{a^{2} + b^{2}}$

$t a n θ = \frac{(a^{2} - b^{2})}{2 a b}$

$c o s e c θ = \frac{a^{2} + b^{2}}{a^{2} - b^{2}}$

$s e c θ = \frac{a^{2} + b^{2}}{2 a b}$
and
$c o t θ = \frac{2 a b}{a^{2} - b^{2}}$

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If sin θ=a2−b2a2+b2, find the values of all T-ratios of θ

Sol: In ΔABC, Let ∠ABC be θ sin θ=a2−b2a2+b2 ⇒ AB = (a2−b2) ⇒ AC = (a2+b2) ⇒ BC = √(a2+b2)2−(a2−b2)2 [According to Pythagoras theorem] ⇒ BC = √4a2b2 ⇒ BC =2ab cosθ=2aba2+b2 tanθ=(a2−b2)2ab cosecθ=a2+b2a2−b2 secθ=a2+b22ab and cotθ=2aba2−b2

If sin $θ = \frac{a^{2} - b^{2}}{a^{2} + b^{2}}$ , find the values of all T-ratios of $θ$