The correct option is B 4 We know that 1 sin^2θ = cos^2θ and 1 + tan^2θ = ^2θ ∴ 4(1 – sin^2θ)(1 + tan^2θ)= 4cos^2θ×^2θ ∵cosθ = 1/sec θ ∴ 4cos^2θ×^2θ = 4 ...

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Standard X

Mathematics

Trigonometric Identities

41-sin 2θ1+ta...

Question

$4 (1 - {sin}^{2} θ) (1 + {tan}^{2} θ) =$

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Solution

The correct option is A
4

We know that $1 - {sin}^{2} θ = {cos}^{2} θ$ and $1 + {tan}^{2} θ = {sec}^{2} θ$
$∴ 4 (1 - {sin}^{2} θ) (1 + {tan}^{2} θ) = 4 {cos}^{2} θ \times {sec}^{2} θ$
$∵ cos θ = \frac{1}{s e c θ}$
$∴ 4 {cos}^{2} θ \times {sec}^{2} θ = 4$

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(1 - {sin}^{2} θ) (1 + {tan}^{2} θ) =

_____.

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Q. If

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, find

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4(1–sin2θ)(1+tan2θ)=

The correct option is A 4 We know that 1−sin2θ=cos2θ and 1+tan2θ=sec2θ ∴4(1–sin2θ)(1+tan2θ)=4cos2θ×sec2θ ∵cosθ=1secθ ∴4cos2θ×sec2θ=4

$4 (1 - {sin}^{2} θ) (1 + {tan}^{2} θ) =$

The correct option is A
4

We know that $1 - {sin}^{2} θ = {cos}^{2} θ$ and $1 + {tan}^{2} θ = {sec}^{2} θ$
$∴ 4 (1 - {sin}^{2} θ) (1 + {tan}^{2} θ) = 4 {cos}^{2} θ \times {sec}^{2} θ$
$∵ cos θ = \frac{1}{s e c θ}$
$∴ 4 {cos}^{2} θ \times {sec}^{2} θ = 4$