If x is an acute angle such that both tan kx and kx is defined for k -0-20- then which of the following is-are correct -A- tan 15 x-tan 11 x-tan 4 x-tan 15 x tan 11 x tan 4 xB- tan 15 x-tan 11 x-tan 4 x-tan 15 x tan 11 x tan 4 xC- 2 x 4 x- 4 x 6 x- 2 x 6 x-1D- 4 x 6 x- 2 x 4 x- 2 x 6 x-1

The correct options are B tan15x tan11x tan4x=tan15xtan11xtan4x nbsp; nbsp; C 2x 4x 4x6x 2x 6x=1 nbsp;tan15x=tan(11x+4x) ⇒tan15x=tan11x+tan4x1 tan11xtan4 ...

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Standard XII

Mathematics

Trigonometric Ratios for Sum of Two Angles

If x is an ac...

Question

If $x$ is an acute angle such that both $tan (k x)$ and $cot (k x)$ is defined for $k \in (0, 20),$ then which of the following is/are correct ?

tan 15 x + tan 11 x + tan 4 x = tan 15 x tan 11 x tan 4 x

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tan 15 x - tan 11 x - tan 4 x = tan 15 x tan 11 x tan 4 x

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cot 2 x cot 4 x - cot 4 x cot 6 x - cot 2 x cot 6 x = 1

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cot 4 x cot 6 x - cot 2 x cot 4 x - cot 2 x cot 6 x = 1

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Solution

The correct options are
B $tan 15 x - tan 11 x - tan 4 x = tan 15 x tan 11 x tan 4 x$
C $cot 2 x cot 4 x - cot 4 x cot 6 x - cot 2 x cot 6 x = 1$
$tan 15 x = tan (11 x + 4 x)$
$\Rightarrow tan 15 x = \frac{tan 11 x + tan 4 x}{1 - tan 11 x tan 4 x}$
$\Rightarrow tan 15 x - tan 15 x tan 11 x tan 4 x = tan 11 x + tan 4 x$
$\Rightarrow tan 15 x - tan 11 x - tan 4 x = tan 15 x tan 11 x tan 4 x$

$cot 6 x = cot (2 x + 4 x)$
$\Rightarrow cot 6 x = \frac{cot 2 x cot 4 x - 1}{cot 2 x + cot 4 x}$
$\Rightarrow cot 6 x cot 2 x + cot 6 x cot 4 x = cot 2 x cot 4 x - 1$
$\Rightarrow cot 2 x cot 4 x - cot 4 x cot 6 x - cot 2 x cot 6 x = 1$

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Trigonometric Ratios for Sum of Two Angles

Standard XII Mathematics

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If x is an acute angle such that both tan(kx) and cot(kx) is defined for k∈(0,20), then which of the following is/are correct ?

If $x$ is an acute angle such that both $tan (k x)$ and $cot (k x)$ is defined for $k \in (0, 20),$ then which of the following is/are correct ?