ot Ao 3 Lec ™ A -tdaun A

1.	ot Ao 3 Lec ™ A -tdaun A
Answer» To prove(CotA -1) /( 2- Sec²A )= CotA / (1+tanA)Solution: LHS = (CotA -1) /( 2- Sec²A ) = ( CotA-1) / (1 + 1 - Sec²A) = (1/tanA)-1 / ( 1 - tan²A) [Since tan²A +1 = sec² So, 1- Sec²A = - tan²A] = (1-tanA) / tanA (1+tanA)(1-tanA) [ identity a² -b² = (a+b) ( a-b)] = 1/ tanA(1+tanA) = CotA / (1+tanA) = RHS Hence Proved Like if you find it useful thanks baby that will be 2*secA not 2-secA u r wrong

Answer»

To prove(CotA -1) /( 2- Sec²A )= CotA / (1+tanA)Solution:

LHS = (CotA -1) /( 2- Sec²A )

= ( CotA-1) / (1 + 1 - Sec²A)

= (1/tanA)-1 / ( 1 - tan²A) [Since tan²A +1 = sec² So, 1- Sec²A = - tan²A]

= (1-tanA) / tanA (1+tanA)(1-tanA) [ identity a² -b² = (a+b) ( a-b)]

= 1/ tanA(1+tanA)

= CotA / (1+tanA) = RHS

Hence Proved

Like if you find it useful

thanks baby

that will be 2*secA

not 2-secA

u r wrong

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