Question

plz write all trignometric formulae on page for class 10th​

Answer 1

Answer:

first of all , for first method we have a simple method to learn this

that is

P B P

H H B

P for perpendicular

B for base

H for hypotenous

and as sin is P/H

cos. B/H

tan. P/B

cosec. H/P

sec.H/B

cot. B/P

Answer 2

Step-by-step explanation:

$\huge{\red{\boxed{\blue{\boxed{\underline{\sf{\red{SOLUTION-}}}}}}}}$

$\huge{\red{\boxed{\green{\boxed{\underline{\sf{\red{FORMULA}}}}}}}}$

>>LET ABC BE A RIGHT ANGLED TRIANGLE IN WHICH / A AND / C ARE ACUTE ANGLES AND AT / B =90°.

>> AC IS HYPOTENUSE

>> BC IS BASE

>> AB IS PERPENDICULAR

>>SO ACCODING TO THIS WE RIGHT SOME BASIC FORMULA OF TRIGONOMETRY.

>> THERE ARE SIX TRIGONOMETRIC RATIOS:

>> SIN θ

>> COS θ

>> TAN θ

>> COT θ

>> COSEC θ

>> SEC θ

>> WHERE (θ) IS ANY ACUTE ANGLE OF RIGHT-ANGLED TRIANGLE.

>> WE TAKE θ AT / A

$\sin(</u><u>A</u><u>) = \frac{p}{h} </u><u>\</u><u>\</u><u> \cos(</u><u>A</u><u>) = \frac{b}{h} \\ \tan(</u><u>A</u><u>) = \frac{p}{b} \\ \cot(</u><u>A</u><u>) = \frac{b}{p} \\ \cosec(a) = \frac{h}{p} \\ \sec(</u><u>A</u><u>) = \frac{b}{h}$

>> ACTUALLY THE MAIN THREE TRIGONOMETRIC RATIOS ARE

>>SIN θ

>>COS θ

>>TAN θ

>> NEXT THREE ARE INVERSE OF THESE LIKE

$=)\sin( A) = \frac{1}{\cosec (A)} \\ =)\cos (A)=\frac{1}{\sec(A )} \\=)\tan (A) = \frac{1}{\cot( A) }$

$\huge{\red{\boxed{\boxed{\underline{\sf{\red{SIMPLE\:FORMULA-}}}}}}}$

$= ) \sin(a) = \frac{1}{ \cosec(a) } \\ = )\cos(a) = \frac{1}{ \sec(a) } \\ = )\tan(a) = \frac{ \sin(a) }{ \cos(a) } \\ = )\cot(a) = \frac{ \cos(a) }{ \sin(a) }$

$\huge{\green{\boxed{\underline{\sf{\red{MAIN\:FORMULA-}}}}}}$

$= )sin(90 - a) = cos(a) \\ = ) \cos(90 - a) = \sin(a) \\ = ) \tan(90 - a) = cot(a) \\ = ) \cot(90 - a) = \tan(a) \\ = ) \sec(90 - a) = \cosec(a) \\ = ) \cosec(90 - a) = \sec(a)$

$\huge{\red{\boxed{\underline{\sf{\red{IDENTITIES-}}}}}}$

$sin^{2}(A)+cos^{2}(A)=1\\1+tan^{2}(A)=sec^{2}(A)\\cot^{2}(A)+1=cosec^{2}(A)$