Question

trigonometry identies​

Answer 1

Answer:1+cosec^2theeta + 1+ sec^2theeta +1+cos^2theeta + 1+sin^2theeta =2

1(cosec^2theeta + sec^2theeta + cos^2theeta + sin^2theeta)

1( 1÷sin^2theeta + 1÷cos^2theeta + 1)

1(1+1). = 2. I think this is wrong but it is right RHS=LHS

Step-by-step explanation:

Answer 2

LHS =

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

lets take two terms at a time and then add

$\frac{1}{1 + \cos^{2} ( \theta ) } + \frac{1}{1 + \sec^{2} ( \theta ) }$

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

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

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

=1

similarly

$\frac{1}{1 + \sin^{2} ( \theta ) } + \frac{1}{1 + \cosec^{2} ( \theta ) }$

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

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

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

=1

adding both

1+ 1 = 2 = RHS