2

Solve equation $1+\sin^2\theta=3\sin\theta\cos\theta \text { (given } \tan71 ^{\circ}34^{\prime}=3) $

My attempt is as follows:-

$$1-2\sin\theta\cos\theta+\sin^2\theta-\sin\theta\cos\theta=0$$ $$\left(\sin\theta-\cos\theta\right)^2+\sin\theta\left(\sin\theta-\cos\theta\right)=0$$ $$\left(\sin\theta-\cos\theta\right)(2\sin\theta-\cos\theta)=0$$ $$\tan\theta=1 \text { or } \tan\theta=\frac{1}{2}$$ $$\theta=n\pi+\frac{\pi}{4}$$

For $\tan\theta=\dfrac{1}{2} \text { how to make use of given condition } \tan71 ^{\circ}34^{\prime}=3$

Bernard
  • 175,478
user3290550
  • 3,452

2 Answers2

2

Use $3=\frac{1+\frac12}{1-1\cdot\frac12}=\tan(45^\circ+\theta)$.

J.G.
  • 115,835
0

As $1+2+3=1\cdot2\cdot3$

$\arctan1+\arctan2+\arctan 3=n\pi$

using Show $\tan(x)+\tan(y)+\tan(z) = \tan(x) \tan(y) \tan(z)$

As $\arctan(u)<\dfrac\pi2$

$n=1$

Now $\arctan1=\dfrac\pi4$

$\arctan2+\arctan\dfrac12=\arctan2+$arccot$2=\dfrac\pi2$

See Are $\mathrm{arccot}(x)$ and $\arctan(1/x)$ the same function?