1.

Prove that: -tan(45 - A)=sec 2A +tan2A​

Answer»

L.H.S.

= 1 / COS(2A) + SIN(2a) / cos(2a)

= (1 + sin(2a)) / (cos(2a))

= (sin^2(a) + cos^2(a) + 2 * sin(a) * cos(a)) / (cos^2(a) - sin^2(a))

= (sin(a) + cos(a))^2 / ((cos(a) - sin(a)) * (cos(a) + sin(a)))

= (cos(a) + sin(a)) / (cos(a) - sin(a))

= cos(a) * (1 + sin(a) / cos(a)) / (cos(a) * (1 - sin(a)/cos(a)))

= (1 + tan(a)) / (1 - tan(a))

= (tan(45) + tan(a)) / (1 - tan(a) * tan(45))

= tan(a + 45)


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