Question

prove that sin theta minus cos theta + 1/ sin theta + cos theta minus one is equal to 1 divided by secant theta minus tan theta

Answer 1

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Answer 2

Answer:

In the question,

We have been provided a trigonometric function,

$\frac{sin\theta-cos\theta+1}{sin\theta+cos\theta-1}=\frac{1}{sec\theta -tan\theta}$

Therefore, on simplifying the given equation by putting in the values of the respective terms in the LHS, we get,

On dividing by 'cosθ' in both Numerator and Denominator, we get,

$\frac{sin\theta-cos\theta+1}{sin\theta+cos\theta-1}=\frac{\frac{sin\theta}{cos\theta}-\frac{cos\theta}{cos\theta}+\frac{1}{cos\theta}}{\frac{sin\theta}{cos\theta}+\frac{cos\theta}{cos\theta}-\frac{1}{cos\theta}}\\=\frac{tan\theta+sec\theta-1}{tan\theta-sec\theta+1}$

Now,

we know that,

sec²θ - tan²θ = 1

On putting in the given equation, we get,

$\frac{tan\theta+sec\theta-1}{tan\theta-sec\theta+1}=\frac{tan\theta+sec\theta-1}{tan\theta-sec\theta+(sec^{2}\theta-tan^{2}\theta)}\\=\frac{tan\theta+sec\theta-1}{(sec\theta-tan\theta)[(sec\theta+tan\theta)-1]}\\=\frac{1}{sec\theta-tan\theta}$

i.e.

$\frac{1}{sec\theta-tan\theta}=\frac{1}{sec\theta-tan\theta}$

Now, we can see here that,

LHS of the equation is equal to the RHS.

Therefore, we can see that the equation has been proved.

Hence, Proved.