{"id":5053,"date":"2026-02-01T19:46:41","date_gmt":"2026-02-01T14:16:41","guid":{"rendered":"https:\/\/physicscatalyst.com\/article\/?p=5053"},"modified":"2026-02-01T19:46:51","modified_gmt":"2026-02-01T14:16:51","slug":"trigonometric-formulas-for-class-10","status":"publish","type":"post","link":"https:\/\/physicscatalyst.com\/article\/trigonometric-formulas-for-class-10\/","title":{"rendered":"Trigonometry formulas for class 10"},"content":{"rendered":"\n

Students can get trigonometry formulas for Class 10 here. Trigonometry is the study of the relationships between triangle angles, lengths, and heights. It contains ratios, functions, identities, and formulas for solving issues based on right-angled triangles. Trigonometry is also used in engineering, astronomy, physics, and architectural design. This chapter is crucial since it covers a wide range of topics such as linear algebra, calculus, and statistics.<\/p>\n\n\n\n

In CBSE Class 10, trigonometry is introduced. It is a completely new and difficult chapter in which one must memorize all of the formulas and apply them correctly. The formulas for Trigonometry Class 10 are listed here.<\/p>\n\n\n\n

Trigonometry formulas List for class 10<\/h2>\n\n\n\n

Trigonometry is quite an important chapter in Class 10 maths and a lot of good questions come in board examination also. Here are the trigonometry formulas for class 10 on one page<\/p>\n\n\n\n

In a right angle triangle ABC where B=90\u00b0<\/p>\n\n\n\n

\"trigonometry<\/a><\/figure>\n\n\n\n

We can define the following term for angle A<\/p>\n\n\n\n

Base<\/strong>:  Side adjacent to the angle<\/p>\n\n\n\n

Perpendicular:<\/strong> Side Opposite of angle<\/p>\n\n\n\n

Hypotenuse<\/strong>: Side opposite to right angle<\/p>\n\n\n\n

We can define the trigonometric ratios for angle A as<\/p>\n\n\n\n

$\\sin A=\u00a0 \\frac {Perpendicular}{Hypotenuse} =\\frac {BC}{AC}$<\/p>\n\n\n\n

$cosec A= \\frac {Hypotenuse}{Perpendicular} =\\frac {AC}{BC}$<\/p>\n\n\n\n

$\\cos A= \\frac {Base}{Hypotenuse} =\\frac {AB}{AC}$<\/p>\n\n\n\n

$sec A= \\frac {Hypotenuse}{Base}=\\frac {AC}{AB}$<\/p>\n\n\n\n

$\\tan A= \\frac {Perpendicular}{Base} =\\frac {BC}{AB}$<\/p>\n\n\n\n

$\\cot A= \\frac {Base}{Perpendicular}=\\frac {AB}{BC}$<\/p>\n\n\n\n

Notice that each ratio in the right-hand column is the inverse, or the reciprocal, of the ratio in the left-hand column.<\/p>\n\n\n\n

Reciprocal of functions<\/strong><\/p>\n\n\n\n

$cosec A = \\frac {1}{\\sin A}$<\/p>\n\n\n\n

$ \\sec A = \\frac {1}{\\cos A}$<\/p>\n\n\n\n

$\\cot A =\\frac {\\cos A }{\\sin A} = \\frac {1}{\\tan A}$<\/p>\n\n\n\n

Trigonometric ratio from complementary angle in Right angle triangle<\/strong><\/p>\n\n\n\n

\"\"<\/a><\/figure>\n\n\n\n

$\\sin C= \\frac {Perpendicular}{Hypotenuse} =\\frac {AB}{AC}$<\/p>\n\n\n\n

$cosec C= \\frac {Hypotenuse}{Perpendicular} =\\frac {AC}{AB}$<\/p>\n\n\n\n

$\\cos C= \\frac {Base}{Hypotenuse} =\\frac {BC}{AC}$<\/p>\n\n\n\n

$\\sec C= \\frac {Hypotenuse}{Base}=\\frac {AC}{BC}$<\/p>\n\n\n\n

$\\tan A= \\frac {Perpendicular}{Base} =\\frac {AB}{BC}$<\/p>\n\n\n\n

$\\cot A= \\frac {Base}{Perpendicular}=\\frac {BC}{AB}$<\/p>\n\n\n\n

This can also be written as C= 90 -A<\/p>\n\n\n\n

$\\sin (90-A) =\\cos(A)$<\/p>\n\n\n\n

$\\cos(90-A) = \\sin A$<\/p>\n\n\n\n

$\\tan(90-A) =\\cot A$<\/p>\n\n\n\n

$\\sec(90-A)= cosec A$<\/p>\n\n\n\n

$cosec (90-A) =\\sec A$<\/p>\n\n\n\n

$\\cot(90- A) =\\tan A$<\/p>\n\n\n\n

Trigonometric identities<\/strong><\/p>\n\n\n\n

$\\sin^2 A + \\cos^2 A =1$
$1 + \\tan^2 A =\\sec^2 A$
$1 + \\cot^2 A =cosec^2 A$<\/p>\n\n\n\n

Trigonometric Ratios of standard angles<\/strong><\/p>\n\n\n\n

\"trigonometric<\/a><\/figure>\n\n\n\n

You can below post on how to remember this easily<\/p>\n\n\n\n

How to easily remember trigonometric ratios table<\/a><\/p>\n\n\n\n

How to Find trigonometric ratios in case one is given<\/strong><\/p>\n\n\n\n

$ \\sin A = \\sqrt {1 – \\cos^2A}$<\/p>\n\n\n\n

$ \\cos A = \\sqrt {1 – \\sin^2A}$<\/p>\n\n\n\n

$ \\tan A = \\frac {\\sin A}{\\sqrt {1 – \\sin^2A}} = \\frac {1}{\\cot A} = \\sqrt { \\sec^2 A -1}\u00a0 = \\frac {1}{\\sqrt { cosec^2 A -1} }$<\/p>\n\n\n\n

$ \\cot A = \\frac {1}{\\tan A}$<\/p>\n\n\n\n

$\\sec A = \\sqrt { 1 + \\tan^2 A} = \\frac {1}{\\cos A}$<\/p>\n\n\n\n

$cosec A = \\sqrt {1 + \\cot^2 A} = \\frac {1}{\\sin A}$<\/p>\n\n\n\n

Range of Values of trigonometric ratios<\/strong><\/p>\n\n\n\n

Values of Sin A and cos A is always less than or equal to 1<\/p>\n\n\n\n

Values of sec A and cosec A is always greater than or equal to 1<\/p>\n\n\n\n

Double Angle Formula<\/strong><\/p>\n\n\n\n

$\\sin 2A = 2 \\sin A \\cos A$
$\\cos 2A= \\cos^2 A – \\sin^2 A =2\\cos ^2 A – 1= 1 – 2 \\sin^2 A= \\frac { 1 – \\tan^2 A}{1+ \\tan^2 A}$
$\\tan 2A= \\frac { 2\\tan A}{1 -\\tan^2 A}$<\/p>\n\n\n\n

Triple Angle Formula<\/strong><\/p>\n\n\n\n

$\\sin 3A = 3 \\sin A – 4 \\sin ^3 A$
$\\cos 3A = 4 \\cos^3 A – 3 \\cos A$
$\\tan 3A= \\frac { 3 \\tan A – \\tan^3 A}{1 – 3\\tan^2 A}$<\/p>\n\n\n\n

Solved Questions<\/h2>\n\n\n\n

Question 1<\/strong><\/p>\n\n\n\n

Find the value of sin A, cos A, and tan A in the triangle having opposite side = 3 cm, adjacent side = 4 cm, hypotenuse = 5 cm.<\/p>\n\n\n\n

Solution<\/strong><\/p>\n\n\n\n