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(1 + tan’0) (1 - cos²0) = tan²0​

User Felix
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5 votes

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)

(1 + tan’0) (1 - cos²0) = tan²0​-example-1
(1 + tan’0) (1 - cos²0) = tan²0​-example-2
(1 + tan’0) (1 - cos²0) = tan²0​-example-3
User N Raghu
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4.8k points