Question

(1 + tan’0) (1 - cos²0) = tan²0​

Answer 1

Answer:

You have to apply Trigonometry Formula :

${sin}^{2} θ + {cos}^{2} θ = 1$

$1 + {tan}^{2} θ = {sec}^{2} θ$

Next, you have to prove it :

$(1 + {tan}^{2} θ)(1 - {cos}^{2} θ) = {tan}^{2} θ$

$LHS = (1 + {tan}^{2} θ)(1 - {cos}^{2} θ)$

$LHS = ( {sec}^{2} θ)( {sin}^{2} θ)$

$LHS = ( \frac{1}{ {cos}^{2} θ} )( {sin}^{2} θ)$

$LHS = \frac{ {sin}^{2} θ}{ {cos}^{2} θ}$

$LHS = {tan}^{2} θ$

$∴LHS = RHS \: (proven)$