Question

prove that cot theta minus cosec theta the whole square equals 1 - cos theta upon 1 + cos theta​

Answer 1

answer :

(cot0-cosec0)

(cos0/sin0-1/sin0)

(cos0-1/sin0)²

(1+cos0)(1-cos0)/(1+cos0)(1-cos0)

1-cos0/1+cos0

hence proved

Answer 2

Answer:

$(cot\theta - cosec \theta)^2$ =  $\frac{(1 - cos\theta )}{(1+ cos\theta)}$ proved.

Step-by-step explanation:

Explanation:

Given that,$(cot\theta - cosec \theta)^2$

And according to the question we need to prove that,

$(cot\theta - cosec \theta)^2 = \frac{1 - cos\theta}{1 + cos\theta}$

So, LHS = $(cot\theta - cosec \theta)^2$ and RHS = $\frac{1 - cos\theta}{1 + cos\theta}$

Step 1:

From the question we have,

LHS = $(cot\theta - cosec \theta)^2$ ..........(i)

As we know, $cot\theta$ = $\frac{cos\theta}{sin\theta}$ and $cosec\theta = \frac{1}{sin\theta}$

Now, we put the value of $cot\theta \ and\ cosec \theta$

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

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

And we also know that $sin^2 \theta = (1 - cos^2\theta)$

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

⇒$\frac{(cos\theta - 1)^2}{(1+ cos\theta)(1 - cos\theta)}$ = $\frac{(cos\theta - 1)(cos\theta - 1)}{-(1+ cos\theta)( cos\theta - 1)}$

⇒$\frac{(cos\theta - 1)}{-(1+ cos\theta)}$ =$\frac{-(cos\theta - 1)}{(1+ cos\theta)}$ = $\frac{(-cos\theta + 1)}{(1+ cos\theta)}$

⇒$\frac{(1 - cos\theta )}{(1+ cos\theta)}$

Here we proved that LHS = RHS = $\frac{(1 - cos\theta )}{(1+ cos\theta)}$

Final answer:

Hence, we proved that $(cot\theta - cosec \theta)^2$ =  $\frac{(1 - cos\theta )}{(1+ cos\theta)}$.

#SPJ2