Prove that \(\dfrac{\sin\theta}{1+\cos\theta}+\dfrac{1+\cos\theta}{\sin\theta}=2\csc\theta\).
\(\displaystyle \dfrac{\sin\theta}{1+\cos\theta}+\dfrac{1+\cos\theta}{\sin\theta}=2\csc\theta\).
Prove that \(\dfrac{\tan A}{1+\sec A}-\dfrac{\tan A}{1-\sec A}=2\csc A\).
\(\displaystyle \dfrac{\tan A}{1+\sec A}-\dfrac{\tan A}{1-\sec A}=2\csc A\).
If \(\tan A=\dfrac{3}{4}\), prove that \(\sin A\cos A=\dfrac{12}{25}\).
\(\displaystyle \sin A\cos A=\dfrac{12}{25}\).
Prove that \((\sin\alpha+\cos\alpha)(\tan\alpha+\cot\alpha)=\sec\alpha+\csc\alpha\).
\(\displaystyle (\sin\alpha+\cos\alpha)(\tan\alpha+\cot\alpha)=\sec\alpha+\csc\alpha\).
Prove that \((\sqrt{3}+1)(3-\cot 30^\circ)=\tan^3 60^\circ-2\sin 60^\circ\).
Both sides equal \(2\sqrt{3}\).
Prove that \(1+\dfrac{\cot^2\alpha}{1+\csc\alpha}=\cosec\alpha\).
\(\displaystyle 1+\dfrac{\cot^2\alpha}{1+\csc\alpha}=\csc\alpha\).
Prove that \(\tan\theta+\tan(90^\circ-\theta)=\sec\theta\,\sec(90^\circ-\theta)\).
Identity holds.
The shadow of a pole of height \(h\) metres is \(\sqrt{3}\,h\) metres long. Find the angle of elevation of the Sun.
\(30^\circ\).
If \(\sqrt{3}\,\tan\theta=1\), find \(\sin^2\theta-\cos^2\theta\).
\(\displaystyle \sin^2\theta-\cos^2\theta=-\dfrac{1}{2}.\)
A \(15\) m long ladder just reaches the top of a vertical wall making an angle of \(60^\circ\) with the wall. Find the height of the wall.
\(7.5\,\text{m}\).
Simplify \((1+\tan^2\theta)(1-\sin\theta)(1+\sin\theta)\).
\(\displaystyle \sec^2\theta\,\cos^2\theta=1-\sin^2\theta=\cos^2\theta\) so the product equals \(1\).
If \(2\sin^2\theta-\cos^2\theta=2\), find \(\theta\).
\(\theta=90^\circ\).
Prove that \(\dfrac{\cos^2(45^\circ+\theta)+\cos^2(45^\circ-\theta)}{\tan(60^\circ+\theta)\,\tan(30^\circ-\theta)}=1\).
The value is \(1\).
An observer \(1.5\) m tall is \(20.5\) m from a tower \(22\) m high. Find the angle of elevation of the top of the tower from the observer's eye.
\(\displaystyle \theta=\tan^{-1}\!\Big(\dfrac{22-1.5}{20.5}\Big)=\tan^{-1}\!\Big(\dfrac{20.5}{20.5}\Big)=\tan^{-1}(1)=45^\circ\).
Prove that \(\tan^4\theta+\tan^2\theta=\sec^4\theta-\sec^2\theta\).
Identity holds true.