Question

cosec A +cot A/ cosec A -cot A =(cosec A+ cot A)^2=1+2cot^2+2cosec A cot A​

Answer 1

Answer:

Proved below.

Step-by-step explanation:

From the properties of trigonometric ratios :

• 1 + cot^2 A = cosec^2 A
• Thus, cosec^2 A - cot^2 A = 1

In the question :

= > ( cosecA + cotA ) / ( cosecA - cotA )

Multiplying denominator as well as numerator by cosecA + cotA:

= > [ ( cosecA + cotA )( cosecA + cotA ) ] / [ ( cosecA - cotA )( cosecA + cotA ) ]

= > ( cosecA + cotA )^2 / ( cosec^2 A - cot^2 A ) { using ( a + b )( a - b ) = a^2 - b^2 }

= > ( cosecA + cotA )^2 { cosec^2 A - cot^2 A = 1 }

= > Proved, for 1.

= > ( cosecA + cotA )^2

= > cosec^2 A + cot^2 A + 2cosecAcotA

= > 1 + cot^2 A + cot^2 A + 2cosecAcotA { cosec^2 A = 1 + cot^2 A }

= > 1 + 2cot^2 A + 2cosecAcotA

Proved.

Answer 2

❚ QuestioN ❚

✩ Prove that ,

$\dfrac{\cosec A+\cot A}{\cosec A-\cot A}=(\cosec A+\cot A)^2\\=1+2\cot^2A+2\cosec A\cot A$

❚ AnsWeR ❚

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✺ 1'st Prove :-

$L.H.S.=\dfrac{\cosec A+\cot A}{\cosec A-\cot A}$

✏ Multiplying the both numerator and denominator by $(\cosec A+\cot A)$

$L.H.S.=\dfrac{(\cosec A+\cot A)(\cosec A+\cot A)}{(\cosec A-\cot A)(\cosec A+\cot A)}$

$L.H.S.=\dfrac{(\cosec A+\cot A)^2}{\cosec^2 A-\cot^2 A}$

$L.H.S.=\dfrac{(\cosec A+\cot A)^2}{1}$

❮ as, $\cosec^2 A-\cot^2 A=1$ ❯

$L.H.S.=(\cosec A+\cot A)^2$

$L.H.S.=R.H.S.\:\red{(proved)}$

✺ 2'nd Prove :-

$L.H.S=(\cosec A+\cot A)^2$

$L.H.S=\cosec^2 A+\cot^2 A+2\cosec A\cot A$

$L.H.S=(1+\cot^2 A)+\cot^2 A+2\cosec A\cot A$

$L.H.S=1+2\cot^2 A+2\cosec A\cot A$

$L.H.S=R.H.S.\:\:\red{(proved)}$

✺ Therefore :-

$\dfrac{\cosec A+\cot A}{\cosec A-\cot A}=(\cosec A+\cot A)^2\\=1+2\cot^2A+2\cosec A\cot A$

(Proved)

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