Question

Ans fast......

step by step explanation...

Don't copy❌

#Dramaqueen⭐

Answer 1

Step-by-step explanation:

LHS:

Given: (CosecA - sinA)(secA - cosA)

$=(\frac{1}{sinA}-sinA)(\frac{1}{cosA}-cosA)$

$=(\frac{1-sin^2A}{sinA})(\frac{1-cos^2A}{cosA})$

$=(\frac{cos^2A}{sinA})(\frac{sin^2A}{cosA} )$

$=cosAsinA$

RHS:

$Given:\frac{1}{tanA+cotA}$

$=\frac{1}{\frac{1}{sinA} + \frac{cosA}{sinA}}$

$=\frac{cosAsinA}{cos^2A+sin^2A}$

$=cosAsinA$

∴ LHS = RHS.

Hope it helps!

Answer 2

QUESTION :

(cosecA-sinA)(secA-cosA) = 1/(tanA+cotA)

ANSWER :

LHS=

(cosecA-sinA)(secA-cosA)

NOW

1/cosA =secA

1/sinA=cosecA

so

LHS

=((1/sinA)-sinA)((1/cosA)-cosA)

= ((1-sin²A)/sinA)((1-cos²A)/cosA)

now as sin²A+cos²A=1

lhs

= (cos²A/sinA)(sin²A/cosA)

= (cos²A.sin²A)/(sinA.cosA)

= sinA.cosA.........1

now RHS=

1/(tanA+cotA)

Now as

tanA=(sinA/cosA)

cotA=(cosA/sinA)

we get

rhs

= 1/((sinA/cosA) + (cosA/sinA))

= 1/((sin²A+cos²A)/sinA.cosA)

but sin²A+cos²A=1

so RHS=

1/(1/sinA.cosA)

= sinA.cosA.......2

hence rhs of 1 and 2 are equal

so LHS will 100% equal

hence

(cosecA-sinA)(secA-cosA) = 1/(tanA+cotA)

hence proved

NOTE :

While solving such questions

we must simplify the lhs as much as possible

but if then also we aren't getting the RHS then simplify RHS too

which will help us to prove the question