Question

tan square theta + cos square theta + 2 is equals to secant squared theta cosec square theta

Answer 1

Answer and Explanation:

To prove : $\tan^2\theta+\cot^2\theta+2=\sec^2\theta \csc^2\theta$

Solution :

Taking LHS,

$=\tan^2\theta+\cot^2\theta+2$

We know, $\tan\theta\cot\theta=1$

So, $=\tan^2\theta+\cot^2\theta+2\tan\theta\cot\theta$

It forming an identity, $a^2+b^2+2ab=(a+b)^2$

$=(\tan\theta+\cot\theta)^2$

$=(\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta})^2$

$=(\frac{\sin^\theta+\cos^\theta}{\cos\theta\sin\theta})^2$

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

$=\frac{1}{\cos^2\theta\sin^2\theta}$

$=\sec^2\theta \csc^2\theta$

=RHS

So, LHS=RHS

Answer 2

Answer:

$tan^{2}\theta+cot^{2}\theta+2=sec^{2}\theta cosec^{2}\theta$

Step-by-step explanation:

$LHS = tan^{2}\theta+cot^{2}\theta+2\\=(1+tan^{2}\theta)+(1+cot^{2}\theta)\\=sec^{2}\theta+cosec^{2}\theta$

\* By Trigonometric identities:

i) 1+tan²A = sec²A

ii) 1+cot²A = cosec²A*/

$=\frac{1}{cos^{2}\theta}+\frac{1}{sin^{2}\theta}\\=\frac{sin^{2}\theta+cos^{2}\theta}{cos^{2}\theta sin^{2}\theta}$

$=\frac{1}{cos^{2}\theta sin^{2}\theta}\\=\frac{1}{cos^{2}\theta}\times \frac{1}{sin^{2}\theta}\\=sec^{2}\theta cosec^{2}\theta \\=RHS$

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