Question

(1+cot square theta)tan theta ÷ sec square theta = cot theta
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Answer 1

• Proof :

L.H.S. = $\frac{(1+cot^{2}\theta)\times tan\theta}{sec^{2}\theta}$

$=\frac{cosec^{2}\theta \times tan\theta}{sec^{2}\theta}$

since $cosec^{2}\theta-cot^{2}\theta=1$

$=\frac{\frac{1}{sin^{2}\theta}\times \frac{sin\theta}{cos\theta}}{\frac{1}{cos^{2}\theta}}$

$=\frac{sin\theta\times cos^{2}\theta}{sin^{2}\theta\times cos\theta}$

$=\frac{cos\theta}{sin\theta}$

$=cot\theta$ = R.H.S.

Hence, proved.

• Trigonometry : It is the study of angles and its sine, cosine, tangent, cosectant, sectant, cotangent ratios.

• Identity Rules :

  >> sinθ * cosecθ = 1

  >> cosθ * secθ = 1

  >> tanθ * cotθ = 1

  >> sin²θ + cos²θ = 1

  >> sec²θ - tan²θ = 1

  >> cosec²θ - cot²θ = 1

Answer 2

Solution:-

Given:-

[(1 + cot²A)/tanA] / (Sec²A)

To Proof :-

[(1 + cot²A)/tanA] / (Sec²A) = cotA

Proof:-

[(1 + cot²A)/tanA] / (Sec²A)

we know that,

( 1+ cot²A) = cosec²A

=) [ cosec²A/ tanA] / sec²A

Converting cosecA , tanA and secA in terms of sinA and cosA.

=) [ (1/sinA)/(sinA/cosA)] / (1/cos²A)

=) [ sinA × cos²A] / [ sin²A × cosA]

=) cosA/sinA

=) cotA

Hence Proved!

Identity Used:-

( 1+ cot²A) = cosec²A

Note:-

cotA = ( cosA/sinA)

tanA = ( sinA/ cosA)

secA = ( 1/ cosA)

cosecA = ( 1/sinA)