Question

From the following figure,
find the values of:
(i) sin A (ii) sec A
(iii) cos² A + sin² A
​

Answer 1

GivEn:

• AB = BC = a

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To find:

• Value of -

1. sin A
2. sec A
3. cos² A + sin² A

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SoluTion:

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${\underline{\sf{\bigstar\;Using\; Pythagoras\; theorem\;:}}}$

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$\;\;\star\;\sf H^2 = B^2 + P^2$

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Therefore,

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$:\implies\sf AC^2 = a^2 + a^2$

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$:\implies\sf AC^2 = 2a^2$

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$:\implies\sf AC^2 = 2a^2$

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$\;\;\;\sf \underline{Taking\;sqrt\;both\;sides\;:}$

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$:\implies\sf \sqrt{AC^2} = \sqrt{2a^2}$

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$:\implies\bf \red{AC = \sqrt{2}a}$

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$\bf We\;have \begin{cases}</p><p> & \text{Base (B) = a} \\</p><p> & \text{Perpendicular (P) = a} \\ & \text{Hypotenuse (H) = \sqrt{2}a }\end{cases}$

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Therefore,

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1) $\sf \sin \;A = \dfrac{P}{H}$

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$:\implies\sf \dfrac{ \cancel{a}}{ \sqrt{2} \cancel{a}}$

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$:\implies\sf \pink{ \dfrac{1}{ \sqrt{2}}}\;\bigstar$

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2) $\sf \sec \;A = \dfrac{H}{B}$

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$:\implies\sf \dfrac{ \sqrt{2} \cancel{a}}{ \cancel{a}}$

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$:\implies\sf \purple{ \sqrt{2}}\;\bigstar$

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3) $\sf cos^2\;A + sin^2 A = \bigg( \dfrac{B}{H} \bigg)^2 + \bigg( \dfrac{P}{H} \bigg)^2$

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$:\implies\sf \bigg( \dfrac{a}{ \sqrt{2}a} \bigg)^2 + \bigg( \dfrac{a}{ \sqrt{2}a} \bigg)^2$

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$:\implies\sf \bigg( \dfrac{ \cancel{a}}{ \sqrt{2} \cancel{a}} \bigg)^2 +\bigg( \dfrac{ \cancel{a}}{ \sqrt{2} \cancel{a}} \bigg)^2$

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$:\implies\sf \bigg( \dfrac{1}{ \sqrt{2}} \bigg)^2 +\bigg( \dfrac{1}{ \sqrt{2}} \bigg)^2$

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$:\implies\sf \dfrac{1}{2} + \dfrac{1}{2}$

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$:\implies\sf \dfrac{1 + 1}{2}$

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$:\implies\sf \cancel{ \dfrac{2}{2}}$

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$:\implies\sf \green{1}\;\bigstar\;\;\;\;\;\;\;\;\;\;\;\;\bigg\lgroup\bf sin^2\;A + cos^2\;A = 1 \bigg\rgroup$

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$\boxed{</p><p>\begin{minipage}{7 cm}</p><p>Fundamental Trigonometric Identities \\ \\</p><p> \sin^2\theta + \cos^2\theta=1 \\ \\</p><p>1+\tan^2\theta = \sec^2\theta \\ \\</p><p>1+\cot^2\theta = \text{cosec}^2 \, \theta </p><p>\end{minipage}</p><p>}$