In ABC, A = 90^(), if C = √3, then find the value of B + C - ^2 B Option 1: 1/2 Option 2: 1/4 Option 3: 0 Option 4: 1/8

Answer: 1/4. Solution: C = √3 C = 60° Since A + B + C = 180°: 90° + B + 60° = 180° B = 30° B + C - ^2 B = 30° + 60° - ^2 30° = 1/2 + 1/2 - (√3/2)^2 = 1 - 3/4 = 1/4

Geometry

Trigonometry

Medium

In $\triangle ABC$ , $\angle A = 90^{\circ}$ , if $\tan C = \sqrt{3}$ , then find the value of $\sin B + \cos C - \cos^2 B$

1/2

1/4

1/8

✅ Correct Option: 2

$\tan C = \sqrt{3} \implies C = 60°$

Since $A + B + C = 180°$ :

$90° + B + 60° = 180°$

$B = 30°$

$\sin B + \cos C - \cos^2 B$

$= \sin 30° + \cos 60° - \cos^2 30°$

$= \dfrac{1}{2} + \dfrac{1}{2} - \left(\dfrac{\sqrt{3}}{2}\right)^2$

$= 1 - \dfrac{3}{4}$

$= \dfrac{1}{4}$

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In △ABC\triangle ABC△ABC, ∠A=90∘\angle A = 90^{\circ}∠A=90∘, if tan⁡C=3\tan C = \sqrt{3}tanC=3​, then find the value of sin⁡B+cos⁡C−cos⁡2B\sin B + \cos C - \cos^2 BsinB+cosC−cos2B

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In $\triangle ABC$ , $\angle A = 90^{\circ}$ , if $\tan C = \sqrt{3}$ , then find the value of $\sin B + \cos C - \cos^2 B$