Solution.

To find the required ratios in this case we use the right angled triangle to find the other side assuming that the opposite side is 3 units while the hypotenuses is 5 units.

To find \(AC\) , we have \begin {align*} AB^2 & = BC^2 + AC^2\\ \implies \quad 5^2 & = 3^2 + AC^2\\ \implies \quad AC^2 & = 25 - 9\\ \implies \quad AC & = \sqrt {16} = 4 \end {align*}

Therefore,

\(\cos \theta = \frac {\text {adj}}{\text {hyp}} = \frac {4}{5}\)

\(\tan \theta = \frac {\text {opp}}{\text {adj}} = \frac {3}{4}\)

Three further ratios are defined as reciprocals of these: cosecant, secant and cotangent.

These are defined as follows

\begin {align*} \text {cosec}\theta & = \frac {1}{\sin \theta }\\\\ \sec \theta & = \frac {1}{\cos \theta }\\\\ \cot \theta & = \frac {1}{\tan \theta }= \frac {\cos \theta }{\sin \theta } \end {align*}

\[\text {From this triangle, we have}\quad \sin \theta = \frac {a}{c}\quad , \quad \cos \theta = \frac {b}{c}\quad \text {and}\quad c^2 = a^2 + b^2\]

\begin {align*} \text {Now},\qquad \sin ^2\theta + \cos ^2\theta & = \left (\sin \theta \right )^2 + \left (\cos \theta \right )^2\\\\ & = \left (\frac {a}{c}\right )^2 + \left (\frac {b}{c}\right )^2\\\\ & = \frac {a^2}{c^2} + \frac {b^2}{c^2}\\\\ & = \frac {a^2 + b^2}{c^2}\\\\ & = \frac {c^2}{c^2}\\\\ \implies \quad \sin ^2\theta + \cos ^2\theta & = 1 \end {align*}

Thus for any angle \(\theta \)

\(*\quad \sin ^2\theta + \cos ^2\theta = 1\)

Proof.

\begin {align*} \text {L.H.S}\qquad \sec ^2 \theta & = \left (\sec \theta \right )^2 = \left (\frac {1}{\cos \theta }\right )^2\\\\ & = \frac {1}{\cos ^2\theta }\\\\ & = \frac {\cos ^2\theta + \sin ^2\theta }{\cos ^2\theta }\\\\ & = \frac {\cos ^2\theta }{\cos ^2\theta } + \frac {\sin ^2\theta }{\cos ^2\theta }\\\\ & = 1 + \tan ^2\theta = \text {R.H.S}\\ \end {align*}

\begin {align*} \text {R.H.S}\qquad 1 + \tan ^2\theta & = 1 + \frac {\sin ^2\theta }{\cos ^2\theta }\\\\ & = \frac {\cos ^2\theta + \sin ^2\theta }{\cos ^2\theta }\\\\ & = \frac {1}{\cos ^2\theta }\\\\ & = \sec ^2\theta = \text {L.H.S}\\ \end {align*} □

Proof.

\begin {align*} \text {L.H.S}\qquad (1 - \cos A)(1 + \sec A) & = 1 (1 + \sec A) - \cos A (1 + \sec A)\\ & = 1 + \sec A - \cos A - \cos A\sec A\\ & = 1 + \frac {1}{\cos A} - \cos A - \cos A\left (\frac {1}{\cos A}\right )\\ & = 1 + \frac {1}{\cos A} - \cos A - 1\\\\ & = \frac {1}{\cos A}-\cos A\\\\ & = \frac {1 - \cos ^2 A}{\cos A}\qquad \left (\sin ^{2}A = 1 - \cos ^{2}A\right )\\\\ & = \frac {\sin ^2A}{\cos A}\\\\ & = \sin A\,\frac {\sin A}{\cos A}\\\\ & = \sin A \tan A = \text {R.H.S}\\\\ \end {align*}

Let \(A\) and \(B\) be two angles. Then \[\sin (A + B) = \sin A \cos B + \cos A \sin B,\] \[\sin (A - B) = \sin A \cos B - \cos A \sin B,\] \[\cos (A + B) = \cos A \cos B - \sin A \sin B,\] \[\cos (A - B) = \cos A \cos B + \sin A \sin B.\] □

Proof.

Note that \(\, \cos 2\theta \,\) and \(\,\cos ^2\theta \,\) are connected by the formula \[\cos 2\theta = \cos ^2\theta - \sin ^2\theta \] But \(\,\sin ^2\theta \,\) doesn’t appear in the given identity. Therefore we must eliminate it.

i.e \(\cos 2\theta = \cos ^2\theta - \sin ^2\theta \,\) But \(\,\sin ^2\theta = 1 - \cos ^2\theta \). Thus \begin {align*} \cos 2\theta & = \cos ^2\theta - ( 1 - \cos ^2\theta )\\ & = \cos ^2\theta - 1 + \cos ^2\theta \\ & = 2\cos ^2\theta - 1 \end {align*}

\[\implies \quad \cos 2\theta + 1 = 2\cos ^2\theta \]

\[\implies \quad \frac {\cos 2\theta + 1}{2} = \cos ^2\theta \] □

Proof. \[\text {L.H.S}\quad \tan (A + B) = \frac {\sin (A+ B)}{\cos (A + B)} = \frac {\sin A\cos B + \sin B \cos A}{\cos A \cos B - \sin A \sin B}\]

Divide both the numerator and the denominator by \(\quad \cos A \cos B\,.\,\) Then

\begin {align*} \tan (A + B) & = \frac {\frac {\sin A \cos B}{\cos A\cos B} + \frac {\sin B\cos A}{\cos A\cos B}}{\frac {\cos A \cos B}{\cos A\cos B} - \frac {\sin A\sin B}{\cos A\cos B}}\\\\ & = \frac {\frac {\sin A}{\cos A} + \frac {\sin B}{\cos B}}{1 - \frac {\sin A}{\cos A}\cdot \frac {\sin B}{\cos B}}\\\\ & = \frac {\tan A + \tan B}{1 - \tan A \tan B}\\\\ \end {align*} □

Proof. \begin {align*} \text {L.H.S},\qquad \sin 2 x & = \frac {\sin 2x}{1}\\\\ & = \frac {2\sin x \cos x}{1}\\\\ & = \frac {2\sin x \cos x}{\cos ^2x + \sin ^2x} \end {align*}

Divide both the numerator and the denominator by \(\, \cos ^2x\,.\,\) Then \begin {align*} \sin 2x & = \frac {\frac {2\sin x \cos x}{\cos ^2x}}{1 + \frac {\sin ^2x}{\cos ^2x}}\\\\ & = \frac {\frac {2\sin x}{\cos x}}{1 + \left (\frac {\sin x}{\cos x}\right )^2}\\\\ & = \frac {2\tan x }{1 + \tan ^2x} = \text {R.H.S}\\\\ \end {align*} □

Solution. (i). Divide \(\sin ^{2}x+\cos ^{2}x=1\) through by \(\sin ^{2}x\): \[1+\frac {\cos ^{2}x}{\sin ^{2}x}=\frac {1}{\sin ^{2}x} \quad \implies \quad 1+\cot ^{2}x=\operatorname {cosec}^{2}x .\]

(ii). From the double angle formula in the form \(\cos 2x=1-2\sin ^{2}x\), \[2\sin ^{2}x=1-\cos 2x \quad \implies \quad \sin ^{2}x=\frac {1-\cos 2x}{2} .\]

(iii). Using the difference formulas and then dividing above and below by \(\cos A\cos B\): \begin {align*} \tan (A-B)&=\frac {\sin (A-B)}{\cos (A-B)} =\frac {\sin A\cos B-\cos A\sin B}{\cos A\cos B+\sin A\sin B}\\[4pt] &=\frac {\frac {\sin A}{\cos A}-\frac {\sin B}{\cos B}} {1+\frac {\sin A\sin B}{\cos A\cos B}} =\frac {\tan A-\tan B}{1+\tan A\tan B} . \end {align*}

(iv). Start from \(\cos 2x=\cos ^{2}x-\sin ^{2}x\) and divide above and below by \(\cos ^{2}x\), which is legitimate wherever \(\cos x\neq 0\): \[\cos 2x=\frac {\cos ^{2}x-\sin ^{2}x}{1} =\frac {\cos ^{2}x-\sin ^{2}x}{\cos ^{2}x+\sin ^{2}x} =\frac {1-\tan ^{2}x}{1+\tan ^{2}x} ,\] replacing the denominator \(1\) by \(\cos ^{2}x+\sin ^{2}x\) first — the standard move for turning an identity into one in \(\tan \) alone.