Trigonometry — NDA Mathematics

NDA trigonometry is CBSE Class 11–12 trigonometry, tested at MCQ speed with a 1/3rd-mark penalty for guessing. There is no calculus of trig functions here and no calculator — just ratios, identities, one addition formula that spawns a dozen corollaries, the general-solution machinery, inverse functions with their fussy domains, and the two classic applications: solving a triangle and measuring a height you cannot climb. Get the addition formula and the elevation-triangle diagram automatic, and most of this section falls into place.

1. What NDA actually asks

Trigonometry carries weightPct 18 in the Mathematics syllabus — second only to Algebra (20%) and ahead of Analytical Geometry (15%) and Calculus (13% + 12%). Against 120 questions at 2.5 marks each, that works out to roughly 21–22 questions worth about 54 marks in a typical paper, spread across five recognisable families:

1. Ratio and identity simplification — given one ratio or one identity, evaluate an expression (often dressed in ${\sin }^2\theta +{\cos }^2\theta =1$ or its secant/cosecant cousins).
2. Compound, multiple and sub-multiple angles — $\sin (A+B)$, $\cos 2A$, $\sin 3A$, half-angle values, or evaluating $\sin 15°$, $\cos 75°$, $\tan 105°$.
3. Trigonometric equations — solve for the general solution, or count solutions in a given interval.
4. Inverse trigonometric functions — principal values, domain/range checks, addition formulas like ${\tan }^{-1}x+{\tan }^{-1}y$.
5. Properties of triangles and heights & distances — sine rule, cosine rule, area, and elevation/depression word problems.

Because wrong answers cost $2.5/3\approx 0.83$ marks, trigonometry punishes guessed sign errors more than it rewards speed — a wrong $\cos (A-B)$ sign or a dropped $\pm$ in a general solution is a two-mark swing (lose the mark you should've earned, pay a penalty on top).

2. The ratio and identity kit

For an acute angle $\theta$ in a right triangle with perpendicular $P$, base $B$, hypotenuse $H$:

$\sin \theta =\frac{P}{H},\cos \theta =\frac{B}{H},\tan \theta =\frac{P}{B},\text{cosec}\theta =\frac{1}{\sin \theta },\sec \theta =\frac{1}{\cos \theta },\cot \theta =\frac{1}{\tan \theta }$

Standard values — own this table cold:

The three Pythagorean identities (all from ${\sin }^2\theta +{\cos }^2\theta =1$, divided through by ${\cos }^2\theta$ or ${\sin }^2\theta$):

${\sin }^2\theta +{\cos }^2\theta =11+{\tan }^2\theta ={\sec }^2\theta 1+{\cot }^2\theta ={\text{cosec}}^2\theta$

Useful rearrangement: $\sec \theta -\tan \theta$ and $\sec \theta +\tan \theta$ are reciprocals, since ${\sec }^2\theta -{\tan }^2\theta =1$. Given one, the other is $1/(\text{that value})$ — a common NDA shortcut.

3. Compound, multiple and sub-multiple angles — the whole kit from one formula

Deriving the addition formula. Take two points on the unit circle at angles $A$ and $B$: $P=(\cos A,\sin A)$, $Q=(\cos B,\sin B)$. The chord $PQ$ subtends angle $A-B$ at the centre, so the distance formula gives $P{Q}^2=2-2\cos (A-B)$. Expanding $P{Q}^2$ directly from coordinates and equating the two expressions yields

$\cos (A-B)=\cos A\cos B+\sin A\sin B.$

Replacing $B$ by $-B$ gives $\cos (A+B)=\cos A\cos B-\sin A\sin B$. Using $\sin \theta =\cos (90°-\theta )$ peels off the sine versions, and dividing sine by cosine gives the tangent versions:

$\sin (A\pm B)=\sin A\cos B\pm \cos A\sin B\cos (A\pm B)=\cos A\cos B\mp \sin A\sin B$ $\tan (A\pm B)=\frac{\tan A\pm \tan B}{1\mp \tan A\tan B}$

Double angle (set $B=A$): $\sin 2A=2\sin A\cos A$; $\cos 2A={\cos }^{2}A-{\sin }^{2}A=2{\cos }^{2}A-1=1-2{\sin }^{2}A$; $\tan 2A=\frac{2\tan A}{1-{\tan }^{2}A}$.

Triple angle (set $B=2A$ in $\sin (A+B)$ and simplify with the double-angle forms): $\sin 3A=3\sin A-4{\sin }^{3}A$; $\cos 3A=4{\cos }^{3}A-3\cos A$; $\tan 3A=\frac{3\tan A-{\tan }^{3}A}{1-3{\tan }^{2}A}$.

Sub-multiple (half-angle) — replace $A$ by $\theta /2$ in the double-angle forms: $\cos \theta =1-2{\sin }^2(\theta /2)=2{\cos }^2(\theta /2)-1$, so $\sin (\theta /2)=\pm \sqrt{\frac{1-\cos \theta }{2}}$ and $\cos (\theta /2)=\pm \sqrt{\frac{1+\cos \theta }{2}}$ (sign fixed by which quadrant $\theta /2$ falls in).

Product-to-sum / sum-to-product (needed for "simplify" questions):

$2\sin A\cos B=\sin (A+B)+\sin (A-B)2\cos A\cos B=\cos (A-B)+\cos (A+B)2\sin A\sin B=\cos (A-B)-\cos (A+B)$ $\sin C+\sin D=2\sin \left(\frac{C+D}{2}\right)\cos \left(\frac{C-D}{2}\right)\cos C+\cos D=2\cos \left(\frac{C+D}{2}\right)\cos \left(\frac{C-D}{2}\right)$

Worked example. Find $\sin 75°$. $\sin 75°=\sin (45°+30°)=\sin 45°\cos 30°+\cos 45°\sin 30°=\frac{\sqrt{2}}{2}\cdot \frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}\cdot \frac{1}{2}=\frac{\sqrt{6}+\sqrt{2}}{4}$.

4. General solutions of trigonometric equations

$\sin \theta =\sin \alpha$ holds at $\theta =\alpha$ and its reflection $\theta =\pi -\alpha$ (both repeating every $2\pi$). Folding these into one family:

$\sin \theta =\sin \alpha \Rightarrow \theta =n\pi +(-1{)}^n\alpha ,n\in \mathbb{Z}$

$\cos \theta =\cos \alpha$ holds at $\theta =\alpha$ and its mirror $\theta =-\alpha$ (both repeating every $2\pi$):

$\cos \theta =\cos \alpha \Rightarrow \theta =2n\pi \pm \alpha ,n\in \mathbb{Z}$

$\tan \theta$ repeats every $\pi$ (not $2\pi$), so:

$\tan \theta =\tan \alpha \Rightarrow \theta =n\pi +\alpha ,n\in \mathbb{Z}$

Special cases worth memorising: $\sin \theta =0\Rightarrow \theta =n\pi$; $\cos \theta =0\Rightarrow \theta =n\pi +\frac{\pi }{2}$; and the squared forms ${\sin }^2\theta ={\sin }^2\alpha$, ${\cos }^2\theta ={\cos }^2\alpha$, ${\tan }^2\theta ={\tan }^2\alpha$ all collapse to the single family $\theta =n\pi \pm \alpha$.

Worked example. Solve $\sin \theta =\frac{1}{2}$: $\alpha =\pi /6$, so $\theta =n\pi +(-1{)}^n\frac{\pi }{6}$. Counting example. How many solutions does $\cos 2\theta =\frac{1}{2}$ have in $[0,2\pi )$? Let $x=2\theta \in [0,4\pi )$. $\cos x=\frac{1}{2}$ gives $x=\frac{\pi }{3},\frac{5\pi }{3},\frac{7\pi }{3},\frac{11\pi }{3}$ — four values in $[0,4\pi )$ — so $\theta$ has 4 solutions in $[0,2\pi )$.

5. Inverse trigonometric functions — domains and ranges

Trig functions aren't one-to-one, so inverses are defined on a restricted (principal-value) branch. NDA questions routinely test whether you know the correct branch:

Complementary-pair identities (each holds on the shared domain):

${\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}{\tan }^{-1}x+{\cot }^{-1}x=\frac{\pi }{2}{\sec }^{-1}x+{\text{cosec}}^{-1}x=\frac{\pi }{2}$

Addition formula (the one NDA loves): for $xy<1$,

${\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\left(\frac{x+y}{1-xy}\right)$

Worked example. ${\tan }^{-1}\frac{1}{2}+{\tan }^{-1}\frac{1}{3}$: here $xy=\frac{1}{6}<1$, so the formula applies directly: $\frac{1/2+1/3}{1-1/6}=\frac{5/6}{5/6}=1\Rightarrow {\tan }^{-1}(1)=\pi /4$. (If $xy>1$ with $x,y>0$, add $\pi$ to the right-hand side before taking ${\tan }^{-1}$ — a frequently-tested exception.)

6. Properties of triangles

For a triangle $ABC$ with sides $a,b,c$ opposite angles $A,B,C$, and circumradius $R$:

Sine rule, derived from the inscribed-angle theorem — drop a diameter from $B$ through the circumcentre to $D$; $\angle BDC=\angle BAC=A$ (same arc), and $\angle BCD=90°$ (angle in a semicircle), so $\sin A=\sin (\angle BDC)=\frac{BC}{BD}=\frac{a}{2R}$:

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$

Cosine rule, derived by coordinates — place $A$ at the origin, $B=(c,0)$, $C=(b\cos A,b\sin A)$; then $a^2=B{C}^2=(b\cos A-c{)}^2+(b\sin A{)}^2=b^2+c^2-2bc\cos A$:

$\cos A=\frac{b^2+c^2-a^2}{2bc}\text{(and cyclically for }\cos B,\cos C\text{)}$

Projection formula: $a=b\cos C+c\cos B$. Area: $\Delta =\frac{1}{2}ab\sin C=\frac{1}{2}bc\sin A=\frac{1}{2}ca\sin B$, and by Heron's formula $\Delta =\sqrt{s(s-a)(s-b)(s-c)}$ where $s=\frac{a+b+c}{2}$.

Worked example. In $△ABC$, $a=5$, $b=7$, $C=60°$. Find $c$. $c^2=25+49-2(5)(7)\cos 60°=74-35=39\Rightarrow c=\sqrt{39}$.

7. Heights and distances

The angle of elevation is measured upward from the horizontal to a point above; the angle of depression is measured downward to a point below — both from the observer's eye. Unless a height is stated for the observer, treat them as a point at ground level. Almost every NDA problem reduces to one or two right triangles sharing the vertical height $h$.

Single triangle: height = (horizontal distance) $\times$ tan(angle of elevation).

Two-position (walking towards the tower): if the elevation is $\alpha$ from a point, and $\beta$ ($>\alpha$) after walking a distance $x$ towards the foot, a reusable formula falls out of two equations $h=d\tan \alpha$ and $h=(d-x)\tan \beta$:

$h=\frac{x\tan \alpha \tan \beta }{\tan \beta -\tan \alpha }$

Worked example. Elevation is 30° from a point; after walking 40 m towards the tower it becomes 60°. $\tan 30°=1/\sqrt{3}$, $\tan 60°=\sqrt{3}$: $h=\frac{40\cdot \frac{1}{\sqrt{3}}\cdot \sqrt{3}}{\sqrt{3}-\frac{1}{\sqrt{3}}}=\frac{40}{\left(\frac{3-1}{\sqrt{3}}\right)}=\frac{40\sqrt{3}}{2}=20\sqrt{3}\text{ m}\approx 34.6\text{ m}.$

Two-tower problems (unequal heights, common base line, elevations from each other's foot or top) and problems needing the sine/cosine rule instead of a right triangle (when the two observation points aren't collinear with the foot) both reduce to the same idea: name the unknown height or distance, write one trig equation per triangle, and solve simultaneously.

8. Solved PYQ-style examples

Q1. If $\tan \theta =\frac{3}{4}$ ($\theta$ acute), find $\sin \theta +\cos \theta$. Solution. 3-4-5 triangle: $\sin \theta =3/5$, $\cos \theta =4/5$. Sum $=7/5$.

Q2. If $\sin \theta +\cos \theta =\frac{7}{5}$, find $\sin \theta \cos \theta$. Solution. Square: $1+2\sin \theta \cos \theta =49/25\Rightarrow \sin \theta \cos \theta =12/25$.

Q3. If $\cos \theta =3/5$ with $0<\theta <\pi /2$, find $\cos 2\theta$. Solution. $\cos 2\theta =2{\cos }^2\theta -1=18/25-1=-7/25$.

Q4. Find the number of solutions of $\cos 2\theta =\frac{1}{2}$ in $[0,2\pi )$. Solution. Worked in Section 4 — 4 solutions.

Q5. Evaluate ${\sin }^{-1}\left(\frac{1}{2}\right)+{\cos }^{-1}\left(\frac{1}{2}\right)$. Solution. By the complementary identity, this is $\pi /2$ directly — no need to even evaluate each term ($\pi /6+\pi /3=\pi /2$ confirms it).

9. Common traps and exam protocol

• Sign confusion in $\cos (A\pm B)$: the cosine addition formula flips signs relative to sine — $\cos (A+B)$ uses a minus, $\sin (A+B)$ uses a plus. Cross-checking with a known value (e.g. $A=B=45°$) catches this in five seconds.
• General solution mix-up: using $n\pi +(-1{)}^n\alpha$ (the sine formula) for a cosine equation, or vice versa. Anchor each to why: sine reflects across the vertical axis (hence $\pm$ folded via $(-1{)}^n$), cosine reflects across the horizontal axis (hence the clean $\pm$ in $2n\pi \pm \alpha$).
• Wrong inverse-trig range: assuming ${\cos }^{-1}x$ shares ${\sin }^{-1}x$'s range $[-\pi /2,\pi /2]$. It doesn't — ${\cos }^{-1}x\in [0,\pi ]$. This single confusion breaks more inverse-trig questions than any formula error.
• ${\tan }^{-1}x+{\tan }^{-1}y$ without checking $xy$ against 1: the plain formula only holds for $xy<1$; for $xy>1$ with both positive, add $\pi$.
• Dropping the $\pm$ in half-angle formulas: $\sin (\theta /2)=\pm \sqrt{(1-\cos \theta )/2}$ needs the sign fixed by the quadrant of $\theta /2$, not assumed positive.
• Elevation/depression triangle errors: mixing up which side is opposite vs adjacent to the given angle, especially in two-triangle "walking" problems — always redraw the triangle and label the angle at the observer's vertex.
• Cosine rule angle-side mismatch: using $\cos A=(b^2+c^2-a^2)/2bc$ but plugging in the wrong side as $a$ — remember $a$ must be the side opposite angle $A$.

Protocol: write the addition formula and the general-solution trio from memory daily for a week — everything else in this chapter derives from those two blocks. Then drill inverse-trig domain/range recall until it's reflexive, and finish with 15–20 mixed heights-and-distances problems, since that's where NDA blends trigonometry with plain triangle-labelling discipline.