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 [3]
If `sin θ = 3/4` prove that `sqrt(("cosec"^2θ - cot^2θ)/(sec^2θ - 1)) = sqrt(7)/3`.
We have `sin theta = 3/4`
In ΔABC
`AC^2 = AB^2 + BC^2`
`=> (4)^2 = (3)^2 + BC^2`
`=> BC^2= 16 - 9`
`=> BC^2 = 7`
`=> BC = sqrt7`
`:. cosec theta = 4/3, sec theta = 4/sqrt7 and cot theta = sqrt7/3`
Now
L.H.S `sqrt((cosec^2 theta - cot^2 theta)/(sec^2 theta - 1))`
`= sqrt(((4/3)^2 - (sqrt7/3)^2)/((4/sqrt7)^2 - 1)`
`= sqrt((16/9 - 7/9)/(16/7 - 1)`
`=sqrt((9/9)/((16 - 7)/7 ))`
`= sqrt(7/9)`
`= sqrt7/3`
= R.H.S
If `tan θ = a/b` then prove that `((a sin theta - b cos theta)/(a sin theta + b cos theta)) = ((a^2 - b^2))/((a^2 + b^2))`.
Let `(a sin theta - b cos theta)/(a sin theta + b cos theta)`
Divide both Nr and Dr with cos θ of (a)
`((a sin theta - b cos theta)/cos theta)/((a sin theta + b cos theta)/cos theta)`
`= (tan theta - b)/(a tan theta + b)`
`=(a xx (a/b) - b)/(a xx (a/b) + b)`
`= (a^2 - b^2)/(a^2 + b^2)`
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.
