Question

If tan (A+B) = 1 and sinA = 1/2 find B​

Answer 1

Given :-

• tan (A + B) = 1
• sin A = 1/2

To find :-

• The value of B

Solution :-

→ tan (A + B) = 1

→ But tan 45° = 1

∴ tan (A + B) = tan 45°

→ $\sf \not tan (A + B) = \not tan 45^{ \circ}$

→ A + B = 45° ------(1)

→ sin A = 1/2

→ But sin 30° = 1/2.

∴ sin A = sin 30°

→ $\sf \not sin A = \not sin 30^{\circ}$

→ A = 30° -------(2)

Substitute (2) in (1)

→ A + B = 45°

→ 30° + B = 45°

→ B = 45° - 30°

→ B = 15°

$\red\bigstar{\boxed{\sf{\pmb{The~~value~~of~~B~~=~~15^{\circ}}}}}$

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Know MorE :-

$\bullet\:\sf Trigonometric\:Values :\\\\\boxed{\begin{tabular}{c|c|c|c|c|c}Radians/Angle & 0 & 30 & 45 & 60 & 90\\\cline{1-6}Sin \theta & 0 & \dfrac{1}{2} & \dfrac{1}{\sqrt{2}} & \dfrac{\sqrt{3}}{2} & 1\\\cline{1-6}Cos \theta & 1 & \dfrac{\sqrt{3}}{2}& \dfrac{1}{\sqrt{2}}& \dfrac{1}{2}&0\\\cline{1-6}Tan \theta&0& \dfrac{1}{\sqrt{3}}&1&\sqrt{3}&Not D{e}fined\end{tabular}}$

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