1+tanA/sinA+1+cotA/cosA=2(secA+cosecA)?

2 Answers
Apr 24, 2018

This should read: Show

# { 1 + tan A}/{sin A} + {1 + cot A}/{cos A} = 2(sec A + csc A)#

Explanation:

I'll assume this is a problem to prove, and should read

Show # { 1 + tan A}/{sin A} + {1 + cot A}/{cos A} = 2(sec A + csc A)#

Let's just get the common denominator and add and see what happens.

# { 1 + tan A}/{sin A} + {1 + cot A}/{cos A} #

# = {cos A(1 + sin A/ cos A) + sin A(1+cos A/sin A)}/{sin A cos A} #

# = {cos A + sin A + sin A + cos A}/{sin A cos A} #

# = {2cos A }/{sin A cos A} + {2 sin A}/{sin A cos A} #

# = 2( 1/sin A + 1/cos A) #

# = 2(csc A + sec A) #

# = 2(sec A + csc A) quad sqrt #

Apr 24, 2018

Verified below

Explanation:

#(1+tanA)/sinA+(1+cotA)/cosA=2(secA+cscA)#

Split the numerator:
#1/sinA+tanA/sinA+1/cosA+cotA/cosA=2(secA+cscA)#

Apply the reciprocal identities: #1/sinA= cscA#, #1/cosA= secA#:
#cscA+tanA/sinA+secA+cotA/cosA=2(secA+cscA)#

Apply the quotient identities: #cotA= cosA/sinA#, #tanA=sinA/cosA#:
#cscA+cancel(sinA)/(cosA/cancel(sinA))+secA+cancel(cosA)/(sinA/cancel(cosA))=2(secA+cscA)#

Apply the reciprocal identities:
#cscA+secA+secA+cscA=2(secA+cscA)#

Combine like terms:
#2cscA+2secA=2(secA+cscA)#

Factor out the 2:
#2(secA+cscA)= 2(secA+cscA)#