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

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

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

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

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

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

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

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

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)`

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.

L.H.S. = sin⁸θ − cos⁸θ

= (sin⁴θ)² – (cos⁴θ)²

= (sin⁴θ – cos⁴θ) (sin⁴θ + cos⁴θ)

= [(sin²θ)² – (cos²θ)²] . [(sin²θ)² + (cos²θ)²] 

= (sin²θ + cos²θ) (sin²θ – cos²θ) . [(sin²θ + cos²θ)² – 2sin²θ.cos²θ] …[∵ a² + b² = (a + b)² – 2ab]

= (1) (sin²θ – cos²θ) (1² – 2sin²θ cos²θ)

= (sin²θ – cos²θ) (1 – 2sin²θ cos²θ)

= R.H.S.

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

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

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

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

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