Question

prove that sin^2A×cosec^2A=1​

Answer 1

AnswEr:

To prove:

• Sin²A × Cosec²A = 1

SoluTion:

We know that,

$\large{\boxed{\sf{\red{Sin^2\:A\:=\:\dfrac{1}{Cosec^2\:A}}}}}$

$\therefore$ $\sf{Sin^2\:\times\:\dfrac{1}{Sin^2\:A}}$

Cancelling Sin²A and Sin²A.

: $\implies$ $\sf{\cancel{Sin^2\:A}\:\times\:\dfrac{1}{\cancel{Sin^2\:A}}}$

: $\implies$ 1

$\large{\therefore}$ $\large{\boxed{\sf{Sin^2\:A\:\times\:Cosec^2\:A\:=1}}}$

Hence proved!

$\rule{200}2$

ExTra InFo:

• Cosec²A - Cot²A = 1

• Sin²A + Cos²A = 1

• Sec²A - Tan²A = 1

Answer 2

$\huge \underline {\underline{ \mathfrak{ \green{AnS}wEr \colon}}}$

We have to prove :

$\large{\boxed{\sf{\sin^2 A \: \times \: \cosec ^2 \: = \: 1}}}$

$\rule{200}{2}$

Take $\underline{ \mathbb{R.H.S}}$

$\implies {\sf{\sin^2 A \: \times \: \cosec ^2 A}} \\ \\ \implies {\sf{\sin^2A \: \times \: \dfrac{1}{\sin^2 A}} \: \: \: \big[\cosec^2A \: = \: \dfrac{1}{\sin^2 A }\big] } \\ \\ \implies {\sf{1}}$

Hence,

$\mathbb{L.H.S \: = \: R.H.S}$

___________________________

Some Identities are :

• Sin²A + Cos²A = 1

• SinA = 1/CosecA

• CosA = 1/SecA

• Sec²A - Tan²A = 1

• Cosec²A - Cot²A = 1

• Sin(90° - A) = Cos A

• Cos(90° - A) = SinA

• Tan(90° - A) = CotA

• Cot(90° - A) = TanA