Hi.Can any one help me with these questions.They are really simple but i cannot seem to get them.Thanks
Q1.Show That:
Q2.Prove that:
Q3.Prove that:
Thanks a lot.
Hi,
I think $\displaystyle Tan(270-\theta)=Cot\theta $
L.H.S
=$\displaystyle tan(270°-x)$
=$\displaystyle \frac{sin(270-\theta}{cos(270-\theta)}$
=$\displaystyle \frac{sin270cos\theta-cos270sin\theta}{cos270cos\theat+sin270sin\theta}$
=$\displaystyle \frac{-cos\theta-0}{0-sin\theta}$
=$\displaystyle \frac{cos\theta}{sin\theta}$
=$\displaystyle cot\theta$
In your Q2
$\displaystyle cos(\alpha+\beta)cos(\alpha-\beta)=cos^2\alpha-sin^2\beta=cos^2\beta-sin^2\alpha$
Take L.H.S
=$\displaystyle \frac{cos(\alpha+\beta)cos(\alpha-\beta)}{1}$
multiply by 2
=$\displaystyle \frac{2cos(\alpha+\beta)cos(\alpha-\beta)}{2}$
=$\displaystyle \frac{cos2\alpha+cos2\beta}{2}$
=$\displaystyle \frac{2cos^2\alpha+2Cos^2\beta-2}{2}$
=$\displaystyle cos^2\alpha+cos^2\beta-1$
=$\displaystyle cos^2\alpha-(1-cos^2\beta)$
=$\displaystyle cos^2\alpha-sin^2\beta$
Or
= $\displaystyle cos^2\beta-(1-cos^2\alpha)$
=$\displaystyle cos^2\beta-sin^2\alpha$
In Your Q3
I think you have a mistake,I think the answer will be $\displaystyle -cos\theta$
It's easier in Q.1 to do
$\displaystyle \displaystyle \begin{align*} \tan{\left(270^{\circ} - \theta\right)} &= \tan{\left(180^{\circ} + 90^{\circ} - \theta\right)} \\ &= \tan{\left(90^{\circ} - \theta\right)} \\ &= \frac{\sin{\left(90^{\circ} - \theta\right)}}{\cos{\left(90^{\circ} - \theta\right)}} \\ &= \frac{\cos{\theta}}{\sin{\theta}} \\ &= \cot{\theta} \end{align*}$