Math, asked by adeepaks2005, 3 months ago

(1+tan2A/1+cot2a) = (1-tanA/1-cotA)2 = tan2A

Answers

Answered by Anonymous
114

Step-by-step explanation:

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  • Prove that

\huge \frac{1 +   \: {tan}^{2} \: A }{1 +   \: {cot}^{2} \: A }  =   { (\frac{1 - tan \: A}{1 - cot \: A} )}^{2}  =  {tan}^{2}  A

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  • We will firstly prove the LHS equal to the RHS and then we will be proving the middle term equal to the RHS.

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\boxed{\begin{minipage}{7cm} Fundamental Trigonometric identities \\ \\ $ \sin^2 \theta + \cos^2 \theta = 1 \\ \\ 1 + \tan^2 \theta = \sec^2 \theta \\ \\ 1 + cot^2 \theta = \text {cosec}^2 \  \theta $ \end{minipage}}

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: ➝ LHS :

\huge \frac{1 +  {tan}^{2} \: A }{1 +  {cot}^{2}  \: A}  =  \frac{1 +  \frac{ {sin}^{2} A}{cos {}^{2} \: A } }{1 +  \frac{ {cos}^{2}A }{ {sin}^{2}A } }

[because we know that tan A = sin A /cos A and cot A = cos A / sin A ]

Now, taking the LCM we have

\huge \frac{ \frac{ {cos}^{2} A \:  +  {sin}^{2}  A}{ {cos}^{2}A } }{ \frac{ {sin}^{2}A +  {cos}^{2}  A}{ {sin}^{2}A } }

Now, as cos^2 A + sin^2 A is common in both the numerator and the denominator, they both will be cancelled out.

Now,

\huge \frac{ \frac{1}{ {cos}^{2} a} }{ \frac{1}{ {sin}^{2} a} }   =  \frac{1  }{ {cos}^{2}A }  \times  \frac{ {sin}^{2}A }{1}  =  {tan}^{2} A

[because sin^2 A / cos^2 A = tan^2 A ]

= RHS

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: ➝ Middle side :

\huge { (\frac{1 - tan \: A}{1 - cot \: A} )}^{2}  = ( \frac{1 -  \frac{sin \: A}{cos \: A} }{1 -  \frac{cos \: A}{sin \: A} } ) {}^{2}

[because we know that tan A = sin A /cos A and cot A = cos A / sin A ]

Taking the LCM we have

\huge =  {( \frac{ \frac{cos \: a - sin \: A}{cos \: A} }{ \frac{sin \: A -  \: cos \: A}{sin \: A} } )} {}^{2}

\huge = ( \frac{cos \: A - sin \: A}{cos \: A}  \times  \frac{sin \: A}{sin \: A - cos \: A} ) {}^{2}

Now, to make sin A - cos A and cos A - sin A equal we will take out -1 and as the whole equation is going to be squared -1 will also be squared:

=  {( - 1)}^{2}  \times  {( \frac{sin \: A - cos \: A}{cos \: A} \times  \frac{sin \: A}{sin \: A - cos \: A}  )}^{2}

 = 1 {( \frac{sin \: A}{cos \: A} )}^{2}  =  {tan}^{2} A

[because sin^2 A / cos^2 A = tan^2 A ].

= RHS

□ Hence, proved.

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