Question

(b) 1+[1-tan theta/1-cot theta]^2
= sec2 theta
​

Answer 1

Answer:

Step-by-step explanation:

1 + [1 - Tanθ/1 - Cotθ]²

= 1 + [1 - Sinθ/Cosθ/1 - Cosθ/Sinθ]²

= 1 + [Cosθ - Sinθ/Cosθ/Sinθ-Cosθ/Sinθ]²

= 1 + Sin²θ/Cos²θ[Cosθ - Sinθ/Sinθ - Cosθ]²

= 1 + Tan²θ[Cos²θ + Sin²θ - 2SinθCosθ/Sin²θ + Cos²θ - 2SinθCosθ]

= 1 + Tan²θ[1 - Sin2θ/ 1 - Sin2θ]

= 1 + Tan²θ

= Sec²θ

= R.H.S.

Hence proved.

Answer 2

Answer:

$LHS = 1+\left(\frac{1-tan\theta}{1-cot\theta}\right)^{2}\\=1+\left(\frac{1-tan\theta}{1-\frac{1}{tan\theta}}\right)^{2}\\=1+\left(\frac{1-tan\theta}{\frac{(tan\theta-1)}{tan\theta}}\right)^{2}\\=1+\left(\frac{tan\theta(1-tan\theta)}{(tan\theta-1)}\right)^{2}\\=1+\left(\frac{-tan\theta(tan\theta-1)}{(tan\theta-1)}\right)^{2}\\=1+(-tan\theta)^{2}\\=1+tan^{2}\theta\\=sec^{2}\theta \\=RHS$

Therefore,

$1+\left(\frac{1-tan\theta}{1-cot\theta}\right)^{2}=sec^{2}\theta$

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