Question

please solve step by step
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Answer 1

Answer:

(b) sec(θ)

Step-by-step explanation:

We have,

$\sf{\dfrac{sin(\theta)\,tan(\theta)}{1-cos(\theta)}+tan^2(\theta)-sec^2(\theta)}$

$\sf{=\dfrac{sin(\theta)\,tan(\theta)}{1-cos(\theta)}-sec^2(\theta)+tan^2(\theta)}$

$\sf{=\dfrac{sin(\theta)\,tan(\theta)}{1-cos(\theta)}-\left\{sec^2(\theta)-tan^2(\theta)\right\}}$

$\sf{=\dfrac{sin(\theta)\,tan(\theta)}{1-cos(\theta)}-1}$

$\sf{=\dfrac{sin(\theta)\cdot\dfrac{sin(\theta)}{cos(\theta)}}{1-cos(\theta)}-1}$

$\sf{=\dfrac{sin^2(\theta)}{cos(\theta)\cdot\left(1-cos(\theta)\right)}-1}$

$\sf{=\dfrac{1-cos^2(\theta)}{cos(\theta)\cdot\left(1-cos(\theta)\right)}-1}$

$\sf{=\dfrac{\left(1-cos(\theta)\right)\left(1+cos(\theta)\right)}{cos(\theta)\cdot\left(1-cos(\theta)\right)}-1}$

$\sf{=\dfrac{1+cos(\theta)}{cos(\theta)}-1}$

$\sf{=\dfrac{1+cos(\theta)-cos(\theta)}{cos(\theta)}}$

$\sf{=\dfrac{1}{cos(\theta)}}$

$\sf{=sec(\theta)}$