Question

prove that1 minus tan theta by 1 minus cot theta the whole square is equal to tan square theta​

Answer 1

Answer:

Answer in attached file

Step-by-step explanation:

Answer 2

Answer with Step-by-step explanation:

Taking

LHS

$(\frac{1-tan\theta}{1-cot\theta})^2$

$\frac{(1-tan\theta)^2}{(1-cot\theta)^2}$

$\frac{(1-\frac{sin\theta}{cos\theta})^2}{(1-\frac{cos\theta}{sin\theta})^2}$

Using formula $tan\theta=\frac{sin\theta}{cos\theta},cot\theta=\frac{cos\theta}{sin\theta}$

$\frac{(\frac{cos\theta-sin\theta}{cos\theta})^2}{(\frac{sin\theta-cos\theta}{sin\theta})^2}$

$\frac{(cos\theta-sin\theta)^2\times sin^2\theta}{(-(cos\theta-sin\theta))^2\times cos^2\theta}$

$\frac{(cos\theta-sin\theta)^2\times sin^2\theta}{(cos\theta-sin\theta)^2\times cos^2\theta}$

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

$tan^2\theta$

LHS=RHS

Hence, proved.

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https://brainly.in/question/14866213:Answered by Ajsa