Question

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Q3. Given that sinθ = a/b, then cosθ is equal to :

A. b/√b² - a²

B. b/a

C. √b² - a²/b

D. a/√b² - a²​​

Answer 1

Answer:

C. (√b^2 - a^2)/b

Step-by-step explanation:

Here's your answer

Answer 2

Given:

$\qquad \implies \sf sin \theta = \dfrac {a}{b}$

To Find:

The value of cosθ = ?

Solution:

Given that, $\sf sin \theta = \dfrac {a}{b}$ Using the identity,

$\implies \sf sin^2 \theta + cos^2 \theta = 1$

$\implies \sf sin^2 A = 1 - cos^2 A$

$\implies \sf sin A = \sqrt{(1 - cos^2 A)}$

Therefore,

$\implies \sf cos \theta = \bigg( \sqrt{ \dfrac {1 - a²}{b^2}} \bigg)$

$\implies \sf \sqrt{(b^2 - a^2)}{b^2}$

$\implies \sf \sqrt{\dfrac {b^2 - a^2}{b}}$

$\large {\underline{\boxed{\sf cos \theta = \sqrt{\dfrac {(b^2 - a^2)}{b}}}}}$

Explore more!!

$\boxed{\begin{minipage}{6cm} Important Trigonometric identities :- \\ \\ \: \: 1)\:\sin^2\theta+\cos^2\theta=1 \\ \\ 2)\:\sin^2\theta= 1-\cos^2\theta \\ \\ 3)\:\cos^2\theta=1-\sin^2\theta \\ \\ 4)\:1+\cot^2\theta=\text{cosec}^2 \, \theta \\ \\5)\: \text{cosec}^2 \, \theta-\cot^2\theta =1 \\ \\ 6)\:\text{cosec}^2 \, \theta= 1+\cot^2\theta \\\ \\ 7)\:\sec^2\theta=1+\tan^2\theta \\ \\ 8)\:\sec^2\theta-\tan^2\theta=1 \\ \\ 9)\:\tan^2\theta=\sec^2\theta-1 \end{minipage}}$

$\large{\underline{\underline{\red{\bf Note:}}}}$

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