Question

cosec A
2+2tana
Cosec A
Cosec A-1
Losec Ati​

Answer 1

$\underline{\bigstar{\sf Given:-}}$

$\bullet\sf \dfrac{CosecA}{CosecA-1}+\dfrac{CosecA}{CosecA+1}= 2+2tan^2A$

$\underline{\bigstar{\sf To \ Prove :-}}$

• We have to Prove the Given Trigonometric Identity !
• Identity used

$\boxed{\sf{\dag\ 1+cot^2A= Cosec^2A}}$

$\boxed{\sf{\dag\ a^2-b^2=(a+b)(a-b)}}$

L.H.S

$\dashrightarrow\sf \dfrac{CosecA}{CosecA-1}+\dfrac{CosecA}{Cosec A+1}\\ \\ \\ \dashrightarrow\sf \dfrac{CosecA(CosecA+1)+ CosecA(Cosec A-1)}{(CosecA-1)(CosecA+1)}\ \ \ \ \big[ By\ taking \ L.C.M \big] \\ \\ \\ \dashrightarrow\sf \dfrac{Cosec^2A+\cancel{CosecA} + Cosec^2A-\cancel{CosecA}}{(CosecA)^2-(1)^2} \ \ \ \ \ \big[ \therefore\ \ (a+b)(a-b)=a^2-b^2\big]\\ \\ \\ \dashrightarrow\sf \dfrac{2Cosec^2A}{\cancel{1}+cot^2A\cancel{-1}}\ \ \ \ \ \big[ \dag\ 1+cot^2A=Cosec^2A\big]\\ \\ \\ \dashrightarrow\sf \dfrac{2(1+cot^2A)}{Cot^2A}\\ \\ \\ \dashrightarrow\sf \dfrac{ 2+2Cot^2A}{Cot^2A}\\ \\ \\ \dashrightarrow\sf \dfrac{2}{Cot^2A}+\dfrac{2\cancel{Cot^2A}}{\cancel{Cot^2A}}\\ \\ \\ \dashrightarrow\sf 2tan^2A+2 \ \ \ \ \Bigg[ \therefore\ \dfrac{1}{Cot A}= tanA\Bigg]\\ \\ \\ \dashrightarrow{\purple{\sf\ 2+2tan^2A = 2+2tan^2A}} \\ \\ \\ \dashrightarrow\sf L.H.S=R.H.S \ \ \ \ (hence\ Proved\ \ !! )$

Answer 2

Answer:

here is your answer

please refer to the attachment

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