Question

1+cot2A/1+cosecA=cosecA

Answer 1

To Find :

$\tt : \implies \dfrac{1 + Cot ^2 A} {1 + Cosec A } =Cosec A.$

SolutioN :

$\tt : \implies \dfrac{1 + Cot ^2 A} {1 + Cosec A }$

★ Apply Formula.

• cot² A = Cosec² A - 1.

$\tt : \implies \dfrac{1 + Cosec ^2 A - 1} {1 + Cosec A }$

★ Apply Formula.

• a² - b² = ( a + b )( a - b )

$\tt : \implies \dfrac{1 + (Cosec A + 1)(Cosec A - 1)} {1 + Cosec A }$

Now,

$\tt : \implies \dfrac{1 + ( \cancel{Cosec A + 1})(Cosec A - 1)} { \cancel{1 + Cosec A} }$

$\tt : \implies 1 + Cosec A - 1$

$\tt : \implies Cosec A.$

LHS = RHS.

Answer 2

Step-by-step explanation:

total resistance when resistors are in triangle patter tell which are in series and which in parallel