Trigonometry is quite a important chapter in Class 10 maths and lot of good questions comes in board examination also. Here is the trigonometric formulas for Class 10 in one page

In a right angle triangle ABC where B=90°

We can define following term for angle A

**Base** : Side adjacent to angle

**Perpendicular:** Side Opposite of angle

**Hypotenuse**: Side opposite to right angle

We can define the trigonometric ratios for angle A as

$sin A= \frac {Perpendicular}{Hypotenuse} =\frac {BC}{AC}$

$cosec A= \frac {Hypotenuse}{Perpendicular} =\frac {AC}{BC}$

$cos A= \frac {Base}{Hypotenuse} =\frac {AB}{AC}$

$sec A= \frac {Hypotenuse}{Base}=\frac {AC}{AB}$

$tan A= \frac {Perpendicular}{Base} =\frac {BC}{AB}$

$cot A= \frac {Base}{Perpendicular}=\frac {AB}{BC}$

Notice that each ratio in the right-hand column is the inverse, or the reciprocal, of the ratio in the left-hand column.

**Reciprocal of functions**

$cosec A = \frac {1}{sin A}$

$ sec A = \frac {1}{cos A}$

$cot A =\frac {cos A }{sin A} = \frac {1}{tan A}$

**Trigonometric ratio from complementary angle in Right angle triangle**

$sin C= \frac {Perpendicular}{Hypotenuse} =\frac {AB}{AC}$

$cosec C= \frac {Hypotenuse}{Perpendicular} =\frac {AC}{AB}$

$cos C= \frac {Base}{Hypotenuse} =\frac {BC}{AC}$

$sec C= \frac {Hypotenuse}{Base}=\frac {AC}{BC}$

$tan A= \frac {Perpendicular}{Base} =\frac {AB}{BC}$

$cot A= \frac {Base}{Perpendicular}=\frac {BC}{AB}$

This can also be written as C= 90 -A

$Sin (90-A) =cos(A)$

$Cos(90-A) = sin A$

$Tan(90-A) =cot A$

$Sec(90-A)= cosec A$

$Cosec (90-A) =sec A$

$Cot(90- A) =tan A$

**Trigonometric identities**

$Sin^2 A + cos^2 A =1$

$1 + tan^2 A =sec^2 A$

$1 + cot^2 A =cosec^2 A$

**Trigonometric Ratios of standard angles**

You can below post on how to remember this easily

How to easily remember trigonometric ratios table

**How to Find trigonometric ratios in case one is given**

$ sin A = \sqrt {1 – cos^2A}$

$ cos A = \sqrt {1 – sin^2A}$

$ tan A = \frac {sin A}{\sqrt {1 – sin^2A}} = \frac {1}{cot A} = \sqrt { sec^2 A -1} = \frac {1}{\sqrt { cosec^2 A -1} }$

$ cot A = \frac {1}{tan A}$

$sec A = \sqrt { 1 + tan^2 A} = \frac {1}{cos A}$

$cosec A = \sqrt {1 + cot^2 A} = \frac {1}{sin A}$

**Range of Values of trigonometric ratios**

Values of Sin A and cos A is always less than or equal to 1

Values of sec A and cosec A is always greater than or equal to 1

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