Question

prove that tan theta + 1 by cos theta whole square + 10 theta minus one by cos theta whole square is equal to 2 into 1 + sin square theta by 1 minus sin square theta

Answer 1

Answer:

Step-by-step explanation:

Proved that,

$(tan\theta+\frac{1}{cas\theta})^{2}+(tan\theta-\frac{1}{cas\theta})^{2}=\frac{2(1+sin^{2}\theta)}{1-sin^{2}\theta}$

RHS

$(tan\theta+\frac{1}{cas\theta})^{2}+(tan\theta-\frac{1}{cas\theta})^{2}\\\\\Rightarrow(\frac{sin\theta}{cos\theta}+\frac{1}{cas\theta})^{2}+(\frac{sin\theta}{cos\theta}-\frac{1}{cas\theta})^{2}\\\\\Rightarrow(\frac{sin\theta+1}{cos\theta})^{2}+(\frac{sin\theta-1}{cas\theta})^{2}\\\\\Rightarrow(\frac{sin^{2}\theta+2sin\theta+1}{cos^{2}\theta})+(\frac{sin^{2}\theta-2sin\theta+1}{cos^{2}\theta})$

$\\\\\Rightarrow(\frac{sin^{2}\theta+2sin\theta+1+sin^{2}\theta-2sin\theta+1}{cos^{2}\theta})\\\\\Rightarrow(\frac{2sin^{2}\theta+2}{cos^{2}\theta})\\\\\Rightarrow\frac{2(sin^{2}\theta+1}{1-sin^{2}\theta})[\text{Because}\text{ }sin^{2}\theta+cos^{2}\theta=1]$

Hence RHS=LHS(Proved)

Answer 2

Answer:

Here's your answer Mate

Step-by-step explanation: