Revision: Trigonometry - 1 Maths HSC Science (General) 11th Standard Maharashtra State Board

\[sineA=\frac{\text{Perpendicular}}{\text{Hypotenuse}}\]

\[cosineA=\frac{\mathrm{Base}}{\text{Hypotenuse}}\]

\[tangentA=\frac{\text{Perpendicular}}{\mathrm{Base}}\]

\[cotangent A = \frac{\text{Base}}{\text{Perpendicular}}\]

\[secantA=\frac{\text{Hypotenuse}}{\mathrm{Base}}\]

\[cosecantA=\frac{\text{Hypotenuse}}{\text{Perpendicular}}\]

L.H.S. = `(tan theta)/(sec theta - 1)`

= `(tan theta)/(sec theta - 1) xx (sec theta + 1)/(sec theta + 1)`

= `(tan theta  (sec theta + 1))/(sec^2 theta - 1)`  ...[a² - b² = (a + b)(a - b)]

= `(tan theta  (sec theta + 1))/(tan^2 theta)  ...[(1 + tan^2 theta = sec^2theta),(tan^2theta = sec^2theta - 1)]`

= `(cancel(tan theta)  (sec theta + 1))/(cancel(tan^2 theta)_(tan theta))`

= `(sec theta + 1)/(tan theta)`

L.H.S. = R.H.S.

Hence proved.

For an acute angle A in a right-angled triangle:

• Hypotenuse is the side opposite the right angle.

• Perpendicular is the side opposite angle A.

• Base is the side adjacent to angle A.

sin(nπ + θ) = (−1)ⁿ sin θ

sin(nπ − θ) = (−1)ⁿ⁻¹ sin θ

cos(nπ ± θ) = (−1)ⁿ cos θ

\[\sin\frac{A}{2}\pm\cos\frac{A}{2}=\pm\sqrt{1\pm\sin A}\]

\[\frac{1-\cos\alpha}{\sin\alpha}=\tan\frac{\alpha}{2},\alpha\neq(2n+1)\pi\]

\[\frac{1+\cos\alpha}{\sin\alpha}=\cot\frac{\alpha}{2},\alpha\neq2n\pi\]

\[\frac{1-\cos\alpha}{1+\cos\alpha}=\tan^{2}\frac{\alpha}{2},\alpha\neq(2n+1)\pi\]