Question

plz solve the given prove that..​

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

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\frac{tan^{3}\theta-1}{tan\theta-1}\\=sec^{2}\theta +tan\theta

Step-by-step explanation:

LHS=\frac{tan^{3}\theta-1}{tan\theta-1}

LHS=\frac{tan^{3}\theta-1^{3}}{tan\theta-1}

/* By algebraic identity:

\boxed {a^{3}-b^{3}=(a-b)(a^{2}+ab+1)} */

=\frac{(tan\theta-1)(tan^{2}\theta+tan\theta\times 1+1^{2})}{(tan\theta-1)}

After cancellation, we get

=tan^{2}\thet+tan\theta\times 1+1^{2}

=(1+tan^{2}\theta)+tan\theta

=sec^{2}\theta +tan\theta

/* By Trigonometric identity:

1+tan²A = sec²A */

=RHS

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