-2 sin-tan- 1 - tan- - 2 sin- sec2- - -1 - tan- 2

Explanation for the correct option:Solving the given expression:2sinθtanθ (1 tanθ )+2sinθ sec^2θ/(1+tanθ )^2=2sinθtanθ 2sinθtan^2θ +2sinθ (1+tan^2θ )/(1+tanθ ) ...

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Byju's Answer

Standard X

Mathematics

Mid Point

[2 sinθtanθ 1...

Question

$\frac{[2 \sin θ \tan θ (1 - \tan θ) + 2 \sin θ s e c^{2} θ]}{{(1 + \tan θ)}^{2}}$

$\frac{\sin θ}{(1 + \tan θ)}$

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$\frac{2 \sin θ}{(1 + \tan θ)}$

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$\frac{2 \sin θ}{{(1 + \tan θ)}^{2}}$

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$N o n e o f t h e s e$

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Solution

The correct option is B
$\frac{2 \sin θ}{(1 + \tan θ)}$

Explanation for the correct option:
Solving the given expression:
$\frac{2 \sin θ \tan θ (1 - \tan θ) + 2 \sin θ s e c^{2} θ }{{(1 + \tan θ)}^{2}}$
$= \frac{2 \sin θ \tan θ - 2 \sin θ \tan^{2} θ + 2 \sin θ (1 + \tan^{2} θ) }{{(1 + \tan θ)}^{2}}$ $[∵ s e c^{2} x = 1 + t a n^{2} x]$
$= \frac{2 \sin θ \tan θ - 2 \sin θ \tan 2 θ + 2 \sin θ + 2 \sin θ \tan^{2} θ }{{(1 + \tan θ)}^{2}}$
$= \frac{2 \sin θ \tan θ + 2 \sin θ }{{(1 + \tan θ)}^{2}}$
$= \frac{2 \sin θ (1 + \tan θ) }{{(1 + \tan θ)}^{2}}$
$= \frac{2 s i n θ }{(1 + t a n θ)}$
Hence, Option ‘B’ is Correct.

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Similar questions

(i) $\frac{s e c θ - 1}{s e c θ + 1} = \frac{s i n^{2} θ}{(1 + c o s θ)^{2}}$

(ii) $\frac{s e c θ - t a n θ}{s e c θ + t a n θ} = \frac{c o s^{2} θ}{(1 + s i n θ)^{2}}$

Prove the following trigonometric identities:

Prove that:

(i) $\sqrt{\frac{s e c θ - 1}{s e c θ + 1}} + \sqrt{\frac{s e c θ + 1}{s e c θ - 1}} = 2 c o s e c θ$

(ii) $\sqrt{\frac{1 + s i n θ}{1 - s i n θ}} + \sqrt{\frac{1 - s i n θ}{1 + s i n θ}} = 2 c o s e c θ$

(iii) $\sqrt{\frac{1 + c o s θ}{1 - c o s θ}} + \sqrt{\frac{1 - c o s θ}{1 + c o s θ}} = 2 c o s e c θ$

(iv) $\frac{s e c θ - 1}{s e c θ + 1} = {(\frac{s i n θ}{1 + c o s θ})}^{2}$

If $\sin (θ_{1} - θ_{2}) = \frac{1}{2}$ and $\cos (θ_{1} + θ_{2}) = \frac{1}{2}, 0 ° < θ_{1} + θ_{2} < 90 °, θ_{1} > θ_{2}$ , then find $θ_{1}$ and $θ_{2}$ .

Q. Prove that:

\frac{1 + t a n^{2} Θ}{1 + c o t^{2} Θ} = (\frac{1 - t a n Θ}{1 - c o t Θ})^{2} = t a n^{2} Θ .

Q. More than One Answer Type
एक से अधिक उत्तर प्रकार के प्रश्न

If

x = (\frac{{sin}^{2} θ {sec}^{2} θ + 1}{{sec}^{2} θ}) + (\frac{{cot}^{2} θ}{1 + c o s e c θ}),

y = \frac{1 + {cos}^{2} θ c o s e c^{2} θ}{c o s e c^{2} θ} + \frac{{tan}^{2} θ}{1 + sec θ}

and

z = \frac{1 - {sin}^{2} θ + sin θ cos θ}{sin θ - {sin}^{3} θ},

where 0° < θ < 90°, then the correct option(s) is/are

यदि

x = (\frac{{sin}^{2} θ {sec}^{2} θ + 1}{{sec}^{2} θ}) + (\frac{{cot}^{2} θ}{1 + c o s e c θ}),

y = \frac{1 + {cos}^{2} θ c o s e c^{2} θ}{c o s e c^{2} θ} + \frac{{tan}^{2} θ}{1 + sec θ}

तथा

z = \frac{1 - {sin}^{2} θ + sin θ cos θ}{sin θ - {sin}^{3} θ},

जहाँ 0° < θ < 90°, तब सही विकल्प है/हैं

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[2sinθtanθ(1-tanθ)+2sinθsec2θ](1+tanθ)2

$\frac{[2 \sin θ \tan θ (1 - \tan θ) + 2 \sin θ s e c^{2} θ]}{{(1 + \tan θ)}^{2}}$