Question

tan 2A / 1+ sec 2A = tan A​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

To prove :-

$\bf \:  \dfrac{tan \: 2 \: A}{1 + sec \: 2 \: A} \: = \: tan \: A$

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$\begin{gathered}\Large{\bold{{\underline{Formula \: Used \::}}}} \end{gathered}$

$\bf \:  ⟼ tan \: A = \dfrac{sin \: A}{cos \: A}$

$\bf \:  ⟼ sin \: 2 \: A \: = \: 2 \: sin \: A \: cos \: A$

$\bf \:  ⟼ \: 1 \: + \: cos \: 2 \: A \: = \: 2 \: {cos}^{2} \: A$

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Proof

Proof Consider LHS :-

$\bf \:  \dfrac{tan \: 2 \: A}{1 + sec \: 2 \: A} \:$

$\sf \:  = \: \dfrac{\dfrac{sin \: 2 \: A}{cos \: 2 \: A} }{1 \: + \: \dfrac{1}{cos \: 2 \: A} }$

$\sf \:  \: = \: \dfrac{sin \: 2 \: A}{ \cancel{cos \: 2 \: A}} \: \times \: \dfrac{ \cancel{cos \: 2 \: A}}{1 \: + \: cos \: 2 \: A}$

$\sf \:  = \: \dfrac{2 \: sin \: A \: cos \: A}{2 \: {cos}^{2} \: A }$

$\sf \:  = \dfrac{sin \: A}{cos \: A}$

$\sf \:  = tan \: A$

$\large{\boxed{\boxed{\bf{Hence, Proved}}}}$

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$\large \: \: \: \: \: \: \: \: \: \: \: {\bf \: ⟼ Explore \: more } ✍$

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Trigonometry Formulas

sin(−θ) = −sin θ

cos(−θ) = cos θ

tan(−θ) = −tan θ

cosec(−θ) = −cosecθ

sec(−θ) = sec θ

cot(−θ) = −cot θ

Product to Sum Formulas

sin x sin y = 1/2 [cos(x–y) − cos(x+y)]

cos x cos y = 1/2[cos(x–y) + cos(x+y)]

sin x cos y = 1/2[sin(x+y) + sin(x−y)]

cos x sin y = 1/2[sin(x+y) – sin(x−y)]

Sum to Product Formulas

sin x + sin y = 2 sin [(x+y)/2] cos [(x-y)/2]

sin x – sin y = 2 cos [(x+y)/2] sin [(x-y)/2]

cos x + cos y = 2 cos [(x+y)/2] cos [(x-y)/2]

cos x – cos y = -2 sin [(x+y)/2] sin [(x-y)/2]

Sum or Difference of angles

cos (A + B) = cos A cos B – sin A sin B

cos (A – B) = cos A cos B + sin A sin B

sin (A+B) = sin A cos B + cos A sin B

sin (A -B) = sin A cos B – cos A sin B

tan(A+B) = [(tan A + tan B)/(1 – tan A tan B)]

tan(A-B) = [(tan A – tan B)/(1 + tan A tan B)]

cot(A+B) = [(cot A cot B − 1)/(cot B + cot A)]

cot(A-B) = [(cot A cot B + 1)/(cot B – cot A)]

cos(A+B) cos(A–B)=cos^2A–sin^2B=cos^2B–sin^2A

sin(A+B) sin(A–B) = sin^2A–sin^2B=cos^2B–cos^2A

Multiple and Submultiple angles

sin2A = 2sinA cosA = [2tan A /(1+tan²A)]

cos2A = cos²A–sin²A = 1–2sin²A = 2cos²A–1= [(1-tan²A)/(1+tan²A)]

tan 2A = (2 tan A)/(1-tan²A)

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