Question

Pls solve this attached trignometry identities problem...

Answer 1

Answer:

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Answer 2

Answer:

1=1 hence proved

Step-by-step explanation:

on cross multiplying we get

sinA/cosA*cosA + 1/cosA*cosA-cosA = tanA-secA+1+sinA/cosA*sinA-tanA+sinA

After cancelling the values, we get

sinA+1-cosA = $sin^{2}$A/cosA+sinA+1-secA

again after cancelling, we get

-cosA+secA = $sin^{2}$A/cosA

(cross multiplying)

-$cos^{2}$A+1/cosA (As we are taking the LCM) = $sin^{2}$A/cosA

bringing the cosA to the other side.

-$cos^{2}$A+1 = $sin^{2}$A/cosA*cosA

After cancelling cosA we get,

-$cos^{2}$A+1 = $sin^{2}$A (Bringing to the other side)

1 = $sin^{2}$A + $cos^{2}$A (Using identity $sin^{2}$A + $cos^{2}$A = 1)

Therefore we finally get,

1=1