Question

please anewer the question attached here ​

Answer 1

Step-by-step explanation:

(1+tan)(1-tan+tan^2)/(1+tan) +tan -sec^2

1-tan +tan^2 +tan -sec^2

0 because

sec^2-tan^2 =1 I can't write theeta

a^3 +b^3 =(a+b)(a^2-ab+b^2)

Answer 2

Answer:

hey here is your solution

pls mark it as brainliest

Step-by-step explanation:

$To \: prove = \\ \frac{1 + tan {}^{3} θ}{1 + tan.θ} + tanθ - sec {}^{2} θ = 0 \\ \\ let \: then \: LHS = \frac{1 + tan {}^{3}θ }{1 + tanθ} + tanθ - sec {}^{2} θ \\ \\ RHS = 0 \\ considering \: LHS \\ = \frac{1 + tan {}^{3} θ}{1 + tanθ} + tanθ - sec {}^{2} θ \\ \\ = \frac{(1) {}^{3} + tan {}^{3} θ}{1 + tanθ} + tanθ - sec {}^{2} θ \\ \\ so \: here \: (1) {}^{3} + tan {}^{3} θ \: is \: in \: a {}^{3} + b {}^{3} \\ so \: we \: know \: that \\ a {}^{3} + b {}^{3} = (a + b)(a {}^{2} + b {}^{2} - ab) \\ \\ hence \: accordingly \\ = \frac{(1 + tan \: θ)(1 + tan {}^{2} θ - tan \: θ)}{1 + tan \:θ } + tan θ - sec {}^{2} θ \\ \\ = 1 + tan {}^{2} θ - tanθ + tanθ - sec {}^{2} θ \\ = sec {}^{2} θ - sec {}^{2} θ \: \: \: \: \: \: \: \: \: \: \: \: \: (1 + tan {}^{2} θ = sec {}^{2} θ) \\ \\ = 0 \\ hence \: LHS=RHS \\ thus \: proved$