Question

if sin x = √99/10 find log sinx +log tanx
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Answer 1

Answer:

≈ -2.

Step-by-step explanation:

To Find:

$\implies \sf log (sinx) + log (tanx)$

Solution:

$\qquad : \implies \sf Sin \: x \: = \: \dfrac{\sqrt{99}}{10}$

$\qquad : \implies \sf log (sin x) + log (tan x)$

$\qquad : \implies \sf log (sin x) + log \bigg( \dfrac{sin x}{cos x} \bigg)$

$\quad : \implies \sf log (sin x) + log (sin x) - log (cos x)$

$\qquad : \implies \sf 2 log (sin x) - log (cos x)$

$\qquad : \implies \sf log (sin^2 x) - log (cos x)$

$\qquad : \implies \sf log (1 - cos^2 x) - log (cos x)$

$\qquad : \implies \sf \bigg( 1 - \dfrac{99}{100} \bigg)$

$\qquad : \implies \sf log \bigg( \dfrac{1}{100} \bigg) - log \bigg( \dfrac{\sqrt{99}}{10} \bigg)$

$\qquad : \implies \sf log 1 - log 100 - log 0.99$

$\qquad : \implies \sf \approx -2$

$\qquad \large{\underline{\boxed{\sf \approx -2}}}$