Question

Please answer the question

Answer 1

tan@−cot@ =tan2 @−cot 2 @

LHS= \frac{tan@-cot@}{[email protected]@} = \frac{tan@- \frac{1}{tan@} }{[email protected]@}

[email protected]@

tan@−cot@

=

[email protected]@

tan@−

tan@

1

= \frac{ tan^{2}@-1 }{[email protected]@.cos@} = \frac{ tan^{2}@-1 }{tan@ \frac{sin@}{cos@} [email protected]@}=

[email protected]@.cos@

tan

2

@−1

=

tan@

cos@

sin@

[email protected]@

tan

2

@−1

= \frac{( tan^{2}@-1) sec^{2}@ }{ tan^{2}@ } = \frac{( tan^{2}@-1)( tan^{2}@+1) }{ tan^{2}@ }=

tan

2

@

(tan

2

@−1)sec

2

@

=

tan

2

@

(tan

2

@−1)(tan

2

@+1)

= \frac{ tan^{4}@-1 }{ tan^{2}@ } = \frac{ tan^{4}@ }{ tan^{2}@ } - \frac{1}{ tan^{2}@ }=

tan

2

@

tan

4

@−1

=

tan

2

@

tan

4

@

−

tan

2

@

1

= tan^{2}@- cot^{2}@ = R.H.s=tan

2

@−cot

2

@=R.H.s

Answer 2

Answer:

Step-by-step explanation:

LHS and RHS is shown in below figure