Question

Sin square a upon cos square a + cos square a upon sin square is equal to 1 by sin square a cos square a minus 2

Answer 1

Answer:

The answer is provided in the attachment.

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Answer 2

Complete question:

Prove that n square an upon cos square a + cos square an upon sin square is equal to 1 by sin square a cos square a minus 2.

Answer:

We proved that, $\frac{sin^2a}{cos^2a} + \frac{cos^2a}{sin^2a}$ = $\frac{1}{sin^2acos^2a} - 2$

Step-by-step explanation:

Explanation:

From the question, we have to prove that,

$\frac{sin^2a}{cos^2a} + \frac{cos^2a}{sin^2a}$ = $\frac{1}{sin^2acos^2a} - 2$

So, we need to prove LHS = RHS.

Step 1:

We have, LHS = $\frac{sin^2a}{cos^2a} + \frac{cos^2a}{sin^2a}$

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

Now, from the question LHS = $\frac{sin^2a}{cos^2a} + \frac{cos^2a}{sin^2a}$

So, this can be written as,

⇒$\frac{sin^2a}{cos^2a} + \frac{cos^2a}{sin^2a} = tan^2a + cot^2a$

And we also know that,

$sec^2\theta = 1 + tan^2\theta$ ⇒ $tan^\theta = sec^2\theta - 1$

and $cosec^2\theta = 1 + cot^2\theta$ ⇒ $cot^2\theta = cosec^2 \theta - 1$

Now, $tan^2a + cot^2a$ can be written as,

⇒$sec^2a - 1 +cosec^2a -1$

⇒$sec^2a + cosce^2a - 2$

⇒ $\frac{1}{cos^2a} + \frac{1}{sin^2a}$ -2

Now we take LCM of $cos^2a \ and\ sin^2a$,

⇒ $\frac{sin^2a +cos^2a}{cos^2asin^2a} -2$ ⇒ $\frac{1}{sin^2acos^2a} -2$

[Where, $sin^2a + cos^2 a = 1$]

So, here we proved that LHS = RHS = $\frac{1}{sin^2acos^2a} -2$

Final answer:

Hence, we proved that, $\frac{sin^2a}{cos^2a} + \frac{cos^2a}{sin^2a}$ = $\frac{1}{sin^2acos^2a} - 2$

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