Question

$show \: that \\ \cos^{2} \alpha + \cos ^{2} \alpha \times \cot^{2} \alpha = \cot^{2} \alpha$
​

Answer 1

Answer:

Given:

${cos}^{2} \alpha + {cos}^{2} \alpha \times {cot }^{2} \alpha = { \cot}^{2} \alpha \\</p><p> { \cos( \alpha ) }^{2} (1 + { \cot( \alpha ) }^{2} ) = { \cot( \alpha ) }^{2} \\ { \cos( \alpha ) }^{2} \times {cosec}^{2} \alpha = { \cot( \alpha ) }^{2} \\ { \cos( \alpha ) }^{2} \times \frac{1}{ { \sin( \alpha ) }^{2} } = { \cot( \alpha ) }^{2} \\ \frac{ { \cos( \alpha ) }^{2} }{ { \sin( \alpha ) }^{2} } = { \cot( \alpha ) }^{2} \\ { \cot( \alpha ) }^{2} = { \cot( \alpha ) }^{2} \\ hence \: proved$

Answer 2

To prove :----

• Cos²A + cos²A*cot²A = cot²A

Solution :-------

→ Cos²A + cos²A*cot²A = cot²A

→ cos²A*cot²A = cot²A - cos²A

Solving RHS now,

we know that, cotA = CosA/sinA

Putting in RHS we get,

→ (cos²A/sin²A) - cos²A

Taking LCM now , we get,

→ (cos²A - cos²A*sin²A)/sin²A

Now we know that, sin²A = (1-cos²A) ,

putting This in Numerator we get,

→ [cos²A - cos²A(1-cos²A)]/sin²A

→ [ cos²A - cos²A + cos⁴A] /sin²A

→ (cos⁴A)/sin²A

→ (cos²A)*(cos²A/sin²A)

→ cos²A*cot²A = LHS .

Hence, Proved ...

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