Question

evaluate the (sec sqare theta) ( cosec square theta- 1)​

Answer 1

Step-by-step explanation:

To prove : (\sec^2\theta -1)(\csc^2\theta-1)=1(sec

2

θ−1)(csc

2

θ−1)=1

Proof :

Taking LHS,

LHS=(\sec^2\theta -1)(\csc^2\theta-1)LHS=(sec

2

θ−1)(csc

2

θ−1)

Using Trigonometric identity,

\sec^2\theta -1=\tan^2\thetasec

2

θ−1=tan

2

θ

\csc^2\theta -1=\cot^2\thetacsc

2

θ−1=cot

2

θ

LHS=(\tan^2\theta)(\cot^2\theta)LHS=(tan

2

θ)(cot

2

θ)

We know, \cot^2\theta=\frac{1}{\tan^2\theta}cot

2

θ=

tan

2

θ

1

LHS=\tan^2\theta\times \frac{1}{\tan^2\theta}LHS=tan

2

θ×

tan

2

θ

1

LHS=1LHS=1

LHS=RHSLHS=RHS

Hence proved.

Answer 2

Answer:

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Step-by-step explanation:

To prove : (\sec^2\theta -1)(\csc^2\theta-1)=1(sec

2

θ−1)(csc

2

θ−1)=1

Proof :

Taking LHS,

LHS=(\sec^2\theta -1)(\csc^2\theta-1)LHS=(sec

2

θ−1)(csc

2

θ−1)

Using Trigonometric identity,

\sec^2\theta -1=\tan^2\thetasec

2

θ−1=tan

2

θ

\csc^2\theta -1=\cot^2\thetacsc

2

θ−1=cot

2

θ

LHS=(\tan^2\theta)(\cot^2\theta)LHS=(tan

2

θ)(cot

2

θ)

We know, \cot^2\theta=\frac{1}{\tan^2\theta}cot

2

θ=

tan

2

θ

LHS=\tan^2\theta\times \frac{1}{\tan^2\theta}LHS=tan

2

θ×

tan

2

θ

1

LHS=1LHS=1