Question

prove it for the brainliest answer ​

Answer 1

Answer:

1- tan³Φ / 1- tanΦ

using a³ - b³ = (a - b)(a²+ ab + b²)

[(1 - tanΦ)(1 + tanΦ + tan²Φ) ]/ (1- tanΦ)

1 + tanΦ + tan²Φ

using identity:- 1 + tan²A = sec²A

sec²Φ + tanΦ

Hence proved!!!!

Answer 2

Answer:

Step-by-step explanation:

we know that $x^{3} - y^{3}$ = (1-x)($x^{2} +y^{2} + xy$)

using the above identity

(1-tanθ)($tan^{2}$θ + 1 + tanθ)/(1-tanθ)

cancelling (1-tanθ) from both numerator from denominator

($tan^{2}$θ + 1 + tanθ)

now we know the trigonometric identity = $tan^{2}$θ + 1 = $sec^{2}$θ

using the above identity, the expression simplifies to

$sec^{2}$θ +tanθ = RHS

Hence Proved

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