Question

If 5 cosA =4, then find the value of (1 – tan2

A)/ sec2

A​

Answer 1

Answer:

5 cosA = 4

so, cosA = 4/5

As, sin2A + cos2A = 1

sin2A = 1 - cos2A

so, sinA = root of ( 1 - cos2A )

= root of [ 1 - (4/5)^2 ]

= root of [1 - ( 16/25) ]

= root of [ (25 - 16 ) / 25 ]

= root of (9/25)

= 3/5

therefore, sinA = 3/5

so, cosA = 4/5 and sinA = 3/5

tanA = sinA/cosA

= [(3/5) / (4/5)]

= (3/5) × (5/4)

= 3/4

tanA = 3/4

Given to find: (1-tan2A)/sec2A

= [1 - (3/4)^2] / sec2A as, tan2A = 1/sec2A

= [1 - (9/16)] / {1÷(3/4)^2}

= [(16 - 9)/16] / (4/3)^2

= (7/16) / (16/9)

= 63/256