Question

show that :tan2@ +tan4@=sec4@_sec2@

Answer 1

Answer:

tan²@+tan⁴@ = sec⁴@-sec²@

Step-by-step explanation:

LHS = tan²@+tan⁴@

= tan²@(1+tan²@)

= tan²@sec²@

/* By Trigonometric identity:

1+tan²A = sec²A */

= (sec²@-1)sec²@

= sec⁴@-sec²@

=RHS

Answer 2

Step-by-step explanation:

We need to show that LHS is equal to RHS of the following trigonometric expression.

$\tan^2\alpha +\tan^4\alpha =\sec^4\alpha -\sec^2\alpha$

Taking LHS of the expression and taking $\tan^2\alpha$ common :

$=\tan^2\alpha +\tan^4\alpha \\\\=\tan^2\alpha (1+\tan^2\alpha )$

Since,

$\sec^2\alpha -\tan^2\alpha =1\\\\\tan^2\alpha =\sec^2\alpha -1$

So,

$=(\sec^2\alpha-1) \sec^2\alpha$

or

$\sec^4\alpha -\sec^2\alpha$

= RHS

So, LHS = RHS

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Trigonometry

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