Question

एक्सप्रेस ऑल ट्रिगोनोमेट्रिक रेश्योस इन द टर्म्स ऑफ़ टेन थीटा​

Answer 1

Step-by-step explanation:

consider @ as thetha pls

sin²@ = 1- cos²@ = 1-1/sec²@1-1/1+tan²@

sin2@= 1 -1/1+tan²@= tan²@/1+tan²@

sin@= tan@/√1+tan²@

similarly for cos@= tan@/√1+tan²@

cosec @= √1+tan²@/tan@

sec@ =√1+tan²@/tan@

cot @= 1/ tan@

hope it may help you

pls mark as brainliests plssss

Answer 2

$\huge \sf { \fcolorbox{green}{g}{✴ \: Explanation \: ✴}}$

Suppose theta = x.

Let tan x= k or. k/1.

Construct a right angled triangle ABC , in which angle B =90°, angle ACB = x and

perpendicular AB= k units , and base BC=1 units.

AC^2= BC^2 + AB^2.

AC^2= (1)^2+(k)^2.

or. Hypotenuse AC= √(1+k^2).

sin x= AB/AC = k/√(1+k^2) = tan x/√(1+tan^2 x).l- .(1).

cos x= BC/AC = 1/√(1+k^2) = 1/√(1+ tan ^2 x)- .(2).

cot x= BC/AB= 1/k = 1/tan x…..(3).

sec x= AC/BC= √(1+k^2)/1. = √(1+tan^2 x)….(4).

cosec x= AC/AB = √(1+k^2)/k = √(1+tan^2 x)/tan x……(5)

_______________________

Or also,

Instead of theeta let us take A

Sin A = tan A/√(1+tan^2 A)

Cos A = 1/√(1+tan^2 A)

Cosec A = {√(1+tan^2 A)}/tan A

Sec A = √(1 + tan^2 A)

CotA = 1/tan A