Question

tan cube theta-1/tan theta-1=sec squared theta+tan theta

Answer 1

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Answer 2

Answer:

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

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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