Formulae [2]
\[sineA=\frac{\text{Perpendicular}}{\text{Hypotenuse}}\]
\[cosineA=\frac{\mathrm{Base}}{\text{Hypotenuse}}\]
\[tangentA=\frac{\text{Perpendicular}}{\mathrm{Base}}\]
\[cotangent A = \frac{\text{Base}}{\text{Perpendicular}}\]
\[secantA=\frac{\text{Hypotenuse}}{\mathrm{Base}}\]
\[cosecantA=\frac{\text{Hypotenuse}}{\text{Perpendicular}}\]
| Sr. No. | Expression | Formulae |
|---|---|---|
| i. | sin (A + B) | sin A cos B + cos A sin B |
| ii. | sin (A − B) | sin A cos B − cos A sin B |
| iii. | cos (A + B) | cos A cos B − sin A sin B |
| iv. | cos (A − B) | cos A cos B + sin A sin B |
| v. | tan (A + B) | \[\frac{\tan A+\tan B}{1-\tan A\tan B}\] |
| vi. | tan (A − B) | \[\frac{\tan A-\tan B}{1+\tan A\tan B}\] |
| vii. | cot (A + B) | \[\frac{\cot A\cot B-1}{\cot A+\cot B}\] |
| viii. | cot (A − B) | \[\frac{\cot A\cot B+1}{\cot B-\cot A}\] |
| ix. | sin(A + B) sin(A − B) |
= sin²A − sin²B |
| x. | cos(A + B) cos(A − B) | = cos²A − sin²B = cos²B − sin²A |
Theorems and Laws [1]
Prove that:
\[\tan x \tan \left( \frac{\pi}{3} - x \right) \tan \left( \frac{\pi}{3} + x \right) = \tan 3x\]
L.H.S = `tanx tan(pi/3 - x) tan (pi/3 + x)`
= `tanx . (sin(pi/3 - x))/(cos(pi/3 - x)).sin(pi/3 + x)/(cos(pi/3 + x))`
= `(sinx . sin(pi/3 - x). sin(pi/3 + x))/(cosx . cos(pi/3 - x) . cos(pi/3 + x))`
= `(sinx . (sin^2 pi/3 - sin^2x))/(cosx . (cos^2 pi/3 - sin^2x))`
= `sinx/cosx((sqrt3/2)^2 - sin^2x)/((1/2)^2 - sin^2x)`
= `sinx/cosx ((3/4) - sin^2x)/((1/4) - sin^2x)`
= `sinx/cosx ((3 - 4sin^2x)/(1 - 4sin^2x))`
= `sinx/cosx ((3 - 4sin^2x)/(1 - 4(1 - cos^2x)))`
= `sinx/cosx ((3 - 4sin^2x)/(4 cos^2x - 3))`
= `(3 sinx - 4 sin^3x)/(4cos^2 - 3cosx)`
= `(sin3x)/(cos3x)`
= `tanx`
Key Points
For an acute angle A in a right-angled triangle:
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Hypotenuse is the side opposite the right angle.
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Perpendicular is the side opposite angle A.
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Base is the side adjacent to angle A.
