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Complementary Trigonometric Ratios

If tan A = co...

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Question 4
If tan A = cot B, prove that $A + B = 90^{\circ}$ .

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Solution

According to question,
tan A = cot B
$\Rightarrow t a n A = t a n (90^{\circ} - B)$
$\Rightarrow A = 90^{\circ} - B$
$\Rightarrow A + B = 90^{\circ}$

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Similar questions

tan A = cot B,

A + B = 90^{\circ} .

A + B = 90^{\circ}

\sqrt{\frac{tan A tan B + tan A cot B}{sin A sec B} - \frac{{sin}^{2} B}{{cos}^{2} A}} = tan A

tan A = cot B

A + B + 90

, where A and B are measures of acute angles.

Q. Question 3
If tan (A + B) =

\sqrt{3}

and tan (A - B) =

\frac{1}{\sqrt{3}}

0^{\circ} < A + B \leq 90^{\circ}; A > B

, find A and B.

tan (A + B) = p, tan (A - B) = q

then prove that

tan 2 A = \frac{p + q}{1 - p q}

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Complementary Trigonometric Ratios

Standard X Mathematics

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