If sin 21- x - y- then 21-sin 69- is equal toA- x 2- y - y 2- x 2B- y 2- x - y 2- x 2C- x 2- y - x 2- y 2D- y 2- x - x 2- y 2

The correct option is A x^2y√(y^2 x^2)sin 21^∘=xy Base =√(y^2 x^2) ⇒ 21^∘=y√(y^2 x^2) ⇒sin 69^∘=cos 21^∘=√(y^2 x^2)y Now, 21^∘ sin 69^∘ =21^∘ cos 21^∘ =y√(y^2 ...

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Byju's Answer

Standard XII

Mathematics

Circular Measurement of Angle

If sin 21∘= x...

Question

If $sin 21^{\circ} = \frac{x}{y}$ , then $sec 21^{\circ} - sin 69^{\circ}$ is equal to

\frac{x^{2}}{y \sqrt{y^{2} - x^{2}}}

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\frac{y^{2}}{x \sqrt{y^{2} - x^{2}}}

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\frac{x^{2}}{y \sqrt{x^{2} - y^{2}}}

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\frac{y^{2}}{x \sqrt{x^{2} - y^{2}}}

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Solution

The correct option is A $\frac{x^{2}}{y \sqrt{y^{2} - x^{2}}}$
$sin 21^{\circ} = \frac{x}{y}$
Base $= \sqrt{y^{2} - x^{2}}$
$\Rightarrow sec 21^{\circ} = \frac{y}{\sqrt{y^{2} - x^{2}}}$
$\Rightarrow sin 69^{\circ} = cos 21^{\circ} = \frac{\sqrt{y^{2} - x^{2}}}{y}$

Now,
$sec 21^{\circ} - sin 69^{\circ}$
$= sec 21^{\circ} - cos 21^{\circ}$
$= \frac{y}{\sqrt{y^{2} - x^{2}}} - \frac{\sqrt{y^{2} - x^{2}}}{y}$
$= \frac{x^{2}}{y \sqrt{y^{2} - x^{2}}}$

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Circular Measurement of Angle

Standard XII Mathematics

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If sin21∘=xy, then sec21∘−sin69∘ is equal to

The correct option is A x2y√y2−x2sin21∘=xy Base =√y2−x2 ⇒sec21∘=y√y2−x2 ⇒sin69∘=cos21∘=√y2−x2y Now, sec21∘−sin69∘ =sec21∘−cos21∘ =y√y2−x2−√y2−x2y =x2y√y2−x2

If $sin 21^{\circ} = \frac{x}{y}$ , then $sec 21^{\circ} - sin 69^{\circ}$ is equal to