Introduction to Trigonometry Class 10: notes and important questions

1. Trigonometric ratios

In a triangle \(ABC\) right-angled at \(B\), for the acute angle \(A\): side opposite \(= BC\), side adjacent \(= AB\), hypotenuse \(= AC\).

• \(\sin A=\dfrac{BC}{AC},\ \cos A=\dfrac{AB}{AC},\ \tan A=\dfrac{BC}{AB}\)
• \(\operatorname{cosec} A=\dfrac{1}{\sin A},\ \sec A=\dfrac{1}{\cos A},\ \cot A=\dfrac{1}{\tan A}\), and \(\tan A=\dfrac{\sin A}{\cos A},\ \cot A=\dfrac{\cos A}{\sin A}\)
• The ratios depend only on the angle, not on the size of the triangle (similar triangles).
• For an acute angle, \(0<\sin A<1\) and \(0<\cos A<1\), because the hypotenuse is the longest side. So \(\sec A>1\) and \(\operatorname{cosec} A>1\).

2. Ratios of specific angles

Memory aid: \(\sin A\) runs \(\sqrt{0}/2,\ \sqrt1/2,\ \sqrt2/2,\ \sqrt3/2,\ \sqrt4/2\); \(\cos A\) runs the same list backwards. \(\operatorname{cosec} 0^\circ\) and \(\cot 0^\circ\) are also not defined. As \(A\) increases from \(0^\circ\) to \(90^\circ\), \(\sin A\) increases from 0 to 1 and \(\cos A\) decreases from 1 to 0.

3. Trigonometric identities

• \(\sin^2A+\cos^2A=1\) (for \(0^\circ\le A\le 90^\circ\))
• \(1+\tan^2A=\sec^2A\) (for \(0^\circ\le A<90^\circ\))
• \(1+\cot^2A=\operatorname{cosec}^2A\) (for \(0^\circ<A\le 90^\circ\))
• Useful factorised forms: \(\sec^2A-\tan^2A=(\sec A-\tan A)(\sec A+\tan A)=1\), so \(\sec A+\tan A=\dfrac{1}{\sec A-\tan A}\). Similarly \(\operatorname{cosec} A+\cot A=\dfrac{1}{\operatorname{cosec} A-\cot A}\).

Worked example 1

If \(4\tan\theta=3\), find \(\dfrac{\sin\theta+\cos\theta}{\sin\theta-\cos\theta}\).

\(\tan\theta=\tfrac34\). Take opposite \(=3k\), adjacent \(=4k\), hypotenuse \(=5k\). Then \(\sin\theta=\tfrac35,\ \cos\theta=\tfrac45\), and the expression \(=\dfrac{7/5}{-1/5}=-7\). (Quicker: divide numerator and denominator by \(\cos\theta\) to get \(\dfrac{\tan\theta+1}{\tan\theta-1}=\dfrac{7/4}{-1/4}=-7\).)

Worked example 2

Evaluate \(\dfrac{3\tan^2 30^\circ+\sec^2 45^\circ}{\operatorname{cosec}^2 60^\circ}\).

\(=\dfrac{3\cdot\frac13+2}{\frac43}=\dfrac{3}{4/3}=\dfrac94\).

Worked example 3

Prove that \(\dfrac{\cos A}{1+\sin A}=\sec A-\tan A\).

RHS \(=\dfrac{1-\sin A}{\cos A}=\dfrac{(1-\sin A)(1+\sin A)}{\cos A(1+\sin A)}=\dfrac{\cos^2A}{\cos A(1+\sin A)}=\dfrac{\cos A}{1+\sin A}=\) LHS.

Common errors

• Writing \(\sin A\) as \(\sin\times A\), or \(\sin^2A\) as \(\sin A^2\).
• Assuming \(\sin(A+B)=\sin A+\sin B\) — check with \(A=B=30^\circ\): \(\sin 60^\circ\ne 1\).
• Mixing up opposite and adjacent sides when the angle is at a different vertex.
• Working on both sides of an identity at once as if it were an equation. Start from one side (usually the more complicated one) and reach the other.
• Forgetting that \(\tan 90^\circ\) and \(\sec 90^\circ\) are not defined.

Board-exam tips

• When one ratio is given, draw a right triangle, use \(k\) for the sides, and find the third side by Pythagoras. Show this step: it earns a mark.
• In identity proofs, convert everything to \(\sin\) and \(\cos\) if you are stuck, and look for \(1-\sin^2A=\cos^2A\) style substitutions.
• Write the values used (for example \(\tan 60^\circ=\sqrt3\)) before simplifying. Examiners give a mark for correct substitution.
• Trigonometric ratios of complementary angles (for example \(\sin(90^\circ-A)=\cos A\)) are not in the current syllabus. You will not be asked to use them.