Question

5 marks question long answer type

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Answer 1

Answer:

$\large \text{ tan^4\theta+tan^2\theta=sec^4\theta-sec^2\theta Shown}$

Explanation:

We have to show

$\large tan^4\theta+tan^2\theta=sec^4\theta-sec^2\theta$

Taking  $\large tan^2 \theta \ and \ sec^2\theta$

common from L.H.S. and R.H.S. respectively we get

$\large tan^2\theta(tan^2\theta+1)=sec^2\theta(sec^2\theta-1)$

Now we have identity

$\large sec^2\theta- tan^2\theta=1$

We can also write it as

$\large sec^2\theta= tan^2\theta+1 \ ... (i)\\\\\\\large tan^2\theta=sec^2\theta-1 \ (ii)$

We have

$\large tan^2\theta(tan^2\theta+1)=sec^2\theta(sec^2\theta-1)\\\\\\\large \text{L.H.S. = tan^2\theta(tan^2\theta+1) From ( i ) and ( ii ) we have }\\\\\\\large \text{L.H.S. = sec^2\theta-1(sec^2\theta) }\\\\\\\large \text{L.H.S. = sec^2\theta(sec^2\theta-1) }$

L.H.S. = R.H.S.

Hence shown.

Answer 2

GIVEN :

tan⁴ø + tan²ø = sec⁴ø - sec²ø

TO PROVE :

tan⁴ø + tan²ø = sec⁴ø - sec²ø

From LHS :

= tan⁴ø + tan²ø

Take out tan²ø as common factor.

= tan²ø ( tan²ø + 1)

= tan²ø (1 + tan²ø)

We know that,

1 + tan²ø = sec²ø

= tan²ø . sec²ø

We know that,

tan²ø = (sec²ø - 1)

= sec²ø - 1 . sec²ø

= (sec²ø - 1)sec²ø

Multiply the terms.

= sec⁴ - sec²ø

Hence, proved !!