If tan A-2 tan B- B- then 2 tan A-B-A- tan BB- BC- 2 cotBD- 2 tan B

The correct option is B cotB 2 tan(A B) = 2( tan A tan B/1 + tanA tan B) = 2 (2 tan B + cot B tan B)/1 + (2 tan B + cot B)tan B = 2 tan B + cot B/2(1 + ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios for Sum of Two Angles

If tan A=2 ta...

Question

If tan A=2tanB+cotB, then 2tan(A-B)=

tanB

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2tanB

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cotB

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2cotB

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Solution

The correct option is C
cotB

2 tan(A - B) = 2 $(\frac{t a n A - t a n B}{1 + t a n A t a n B})$

= 2 $\frac{(2 t a n B + c o t B - t a n B)}{1 + (2 t a n B + c o t B) t a n B}$ = 2 $\frac{t a n B + c o t B}{2 (1 + t a n^{2} B)}$

= $\frac{c o t B (t a n^{2} B + 1)}{(1 + t a n^{2} B)}$ = cot B.

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If tan A=2tanB+cotB, then 2tan(A-B)=

The correct option is C cotB 2 tan(A - B) = 2(tanA−tanB1+tanAtanB) = 2 (2tanB+cotB−tanB)1+(2tanB+cotB)tanB = 2 tanB+cotB2(1+tan2B) = cotB(tan2B+1)(1+tan2B) = cot B.

The correct option is C
cotB

2 tan(A - B) = 2 $(\frac{t a n A - t a n B}{1 + t a n A t a n B})$

= 2 $\frac{(2 t a n B + c o t B - t a n B)}{1 + (2 t a n B + c o t B) t a n B}$ = 2 $\frac{t a n B + c o t B}{2 (1 + t a n^{2} B)}$

= $\frac{c o t B (t a n^{2} B + 1)}{(1 + t a n^{2} B)}$ = cot B.