Easy Practice — Set 1

Quick Recap — Trigonometric Ratios

• Standard values: $\sin30^\circ=\tfrac12$, $\sin45^\circ=\tfrac1{\sqrt2}$, $\sin60^\circ=\tfrac{\sqrt3}{2}$; $\cos$ mirrors these.
• Identities: $\sin^2\theta+\cos^2\theta=1$, $1+\tan^2\theta=\sec^2\theta$, $1+\cot^2\theta=\csc^2\theta$.
• Reciprocals: $\sec=\tfrac1{\cos}$, $\csc=\tfrac1{\sin}$, $\cot=\tfrac1{\tan}$.
• $\sin$ is odd, $\cos$ is even; $\sin$ and $\cos$ have period $2\pi$ and range $[-1,1]$.

Beyond-NCERT JEE Formulae

Inverse-trig sum and difference. For arctangents,

$$\tan^{-1}x+\tan^{-1}y=\tan^{-1}\dfrac{x+y}{1-xy}.$$

This is exact when $xy<1$; if $xy>1$, add $\pi$ when $x>0$ and subtract $\pi$ when $x<0$. The double-angle chain is

$$2\tan^{-1}x=\tan^{-1}\dfrac{2x}{1-x^2}=\sin^{-1}\dfrac{2x}{1+x^2}=\cos^{-1}\dfrac{1-x^2}{1+x^2},$$

valid for $0\le x\le 1$ (the $\tan^{-1}$ piece needs $x<1$, the $\cos^{-1}$ piece needs $x\ge 0$). The sine and cosine analogues are

$$\sin^{-1}x\pm\sin^{-1}y=\sin^{-1}\!\left(x\sqrt{1-y^2}\pm y\sqrt{1-x^2}\right),$$

$$\cos^{-1}x\pm\cos^{-1}y=\cos^{-1}\!\left(xy\mp\sqrt{1-x^2}\sqrt{1-y^2}\right).$$

[JEE Tip] Each of these holds only on a restricted domain: after applying a formula, confirm the answer lies in the principal range, adding or subtracting $\pi$ (or flipping a sign) when it does not.

Product-to-sum.

$$2\sin A\cos B=\sin(A+B)+\sin(A-B),$$

$$2\cos A\cos B=\cos(A-B)+\cos(A+B),$$

$$2\sin A\sin B=\cos(A-B)-\cos(A+B).$$

Sum-to-product.

$$\sin C+\sin D=2\sin\dfrac{C+D}{2}\cos\dfrac{C-D}{2},\qquad \sin C-\sin D=2\cos\dfrac{C+D}{2}\sin\dfrac{C-D}{2},$$

$$\cos C+\cos D=2\cos\dfrac{C+D}{2}\cos\dfrac{C-D}{2},\qquad \cos C-\cos D=-2\sin\dfrac{C+D}{2}\sin\dfrac{C-D}{2}.$$

Conditional identities (for $A+B+C=\pi$).

$$\tan A+\tan B+\tan C=\tan A\tan B\tan C,$$

$$\sin 2A+\sin 2B+\sin 2C=4\sin A\sin B\sin C,$$

$$\cos A+\cos B+\cos C=1+4\sin\tfrac{A}{2}\sin\tfrac{B}{2}\sin\tfrac{C}{2}.$$

The half-angles obey $\tfrac{A}{2}+\tfrac{B}{2}+\tfrac{C}{2}=\tfrac{\pi}{2}$, so they also satisfy $\cot\tfrac{A}{2}+\cot\tfrac{B}{2}+\cot\tfrac{C}{2}=\cot\tfrac{A}{2}\cot\tfrac{B}{2}\cot\tfrac{C}{2}$.

Triangle relations. The projection formula reads $a=b\cos C+c\cos B$ (with the companions for $b$ and $c$ by cyclic rotation). Writing $\Delta$ for the area and $s=\tfrac{a+b+c}{2}$ for the semi-perimeter, the circumradius $R$ and inradius $r$ are

$$R=\dfrac{abc}{4\Delta},\qquad r=\dfrac{\Delta}{s}.$$

[JEE Tip] These lock onto the cosine-sum identity through $\cos A+\cos B+\cos C=1+\dfrac{r}{R}$, converting a sum of cosines straight into a radius ratio.

Solved Examples — Beyond-NCERT Formulae

Each example applies one of the Beyond-NCERT tools above; several carry a second-method check.

Example 1 — Arctangent sum, no correction

Evaluate $\tan^{-1}\tfrac{1}{2}+\tan^{-1}\tfrac{1}{3}$. Here $xy=\tfrac{1}{2}\cdot\tfrac{1}{3}=\tfrac{1}{6}<1$, so the plain rule applies:

$$\tan^{-1}\tfrac{1}{2}+\tan^{-1}\tfrac{1}{3}=\tan^{-1}\dfrac{\tfrac{1}{2}+\tfrac{1}{3}}{1-\tfrac{1}{6}}=\tan^{-1}\dfrac{5/6}{5/6}=\tan^{-1}1=\dfrac{\pi}{4}.$$

Example 2 — A $2\tan^{-1}$ conversion

Express $2\tan^{-1}\tfrac{1}{3}$ as a single inverse function. With $x=\tfrac{1}{3}$,

$$\dfrac{2x}{1-x^2}=\dfrac{2/3}{8/9}=\dfrac{3}{4},\qquad \dfrac{2x}{1+x^2}=\dfrac{2/3}{10/9}=\dfrac{3}{5},\qquad \dfrac{1-x^2}{1+x^2}=\dfrac{8/9}{10/9}=\dfrac{4}{5}.$$

Since $0<x<1$ all three branches describe the same acute angle, so $2\tan^{-1}\tfrac{1}{3}=\tan^{-1}\tfrac{3}{4}=\sin^{-1}\tfrac{3}{5}=\cos^{-1}\tfrac{4}{5}\approx 36.87^\circ$.

Example 3 — Arcsine addition

Prove $\sin^{-1}\tfrac{3}{5}+\sin^{-1}\tfrac{5}{13}=\sin^{-1}\tfrac{56}{65}$. Take $x=\tfrac{3}{5}$ so $\sqrt{1-x^2}=\tfrac{4}{5}$, and $y=\tfrac{5}{13}$ so $\sqrt{1-y^2}=\tfrac{12}{13}$. Then

$$x\sqrt{1-y^2}+y\sqrt{1-x^2}=\dfrac{3}{5}\cdot\dfrac{12}{13}+\dfrac{5}{13}\cdot\dfrac{4}{5}=\dfrac{36}{65}+\dfrac{20}{65}=\dfrac{56}{65}.$$

Because $x^2+y^2=\tfrac{2146}{4225}<1$, the result stays within the principal range, so the identity holds as written.

Example 4 — Product-to-sum

Evaluate $2\cos 75^\circ\cos 15^\circ$. The rule $2\cos A\cos B=\cos(A-B)+\cos(A+B)$ gives

$$2\cos 75^\circ\cos 15^\circ=\cos 60^\circ+\cos 90^\circ=\dfrac{1}{2}+0=\dfrac{1}{2}.$$

Example 5 — Sum-to-product simplification

Show $\cos 20^\circ+\cos 100^\circ+\cos 140^\circ=0$. Combine the first and third terms, then the survivor with the middle term:

$$\cos 20^\circ+\cos 140^\circ=2\cos 80^\circ\cos 60^\circ=\cos 80^\circ,\qquad \cos 80^\circ+\cos 100^\circ=2\cos 90^\circ\cos 10^\circ=0.$$

Since $\cos 90^\circ=0$, the entire sum collapses to $0$.

Example 6 — Sum-to-product to solve an equation

Solve $\cos\theta+\cos 2\theta+\cos 3\theta=0$. Pairing the outer terms, $\cos\theta+\cos 3\theta=2\cos 2\theta\cos\theta$, so the equation factors:

$$\cos 2\theta\,(2\cos\theta+1)=0.$$

Thus $\cos 2\theta=0\Rightarrow\theta=(2n+1)\dfrac{\pi}{4}$, or $\cos\theta=-\dfrac{1}{2}\Rightarrow\theta=2n\pi\pm\dfrac{2\pi}{3}$, with $n\in\mathbb{Z}$.

Example 7 — A conditional identity on a fixed triangle

For any triangle $A+B+C=\pi$, so $\tan C=-\tan(A+B)=-\dfrac{\tan A+\tan B}{1-\tan A\tan B}$; clearing the denominator rearranges to $\tan A+\tan B+\tan C=\tan A\tan B\tan C$. Test it on $A=45^\circ,\,B=60^\circ,\,C=75^\circ$ with $\tan 75^\circ=2+\sqrt{3}$:

$$\tan A+\tan B+\tan C=1+\sqrt{3}+(2+\sqrt{3})=3+2\sqrt{3},$$

$$\tan A\tan B\tan C=1\cdot\sqrt{3}\cdot(2+\sqrt{3})=2\sqrt{3}+3,$$

and the two sides match.

Example 8 — Circumradius and inradius from $R=\dfrac{abc}{4\Delta}$

Find $R$ and $r$ for the right triangle with sides $6,8,10$. The legs $6$ and $8$ meet at the right angle, so $\Delta=\tfrac{1}{2}\cdot 6\cdot 8=24$ and $s=\tfrac{6+8+10}{2}=12$. Then

$$R=\dfrac{abc}{4\Delta}=\dfrac{6\cdot 8\cdot 10}{4\cdot 24}=5,\qquad r=\dfrac{\Delta}{s}=\dfrac{24}{12}=2.$$

Both agree with the right-triangle shortcuts $R=\tfrac{\text{hypotenuse}}{2}=5$ and $r=\tfrac{6+8-10}{2}=2$.