Question

can any body help me to do this question​

Answer 1

QUESTION:

$If \:\sin(\theta) = 12 \: find \\ (i) \sec(\theta) - \tan(\theta) \\ (ii) \frac{1}{ {\cos}^{2}\theta } - { \tan }^{2} \theta$

ANSWER:

$\sec(\theta) - \tan(\theta) = \frac{11}{ \sqrt{143} }\\ \frac{1}{ {\cos}^{2}\theta } - { \tan }^{2} \theta = 1$

GIVEN:

$\sin(\theta) = 12$

FORMULA:

$\cos(\theta) = \sqrt{1 - { \sin}^{2} \theta} \\ \tan(\theta) = \frac{ \sin(\theta) }{ \cos(\theta) } \\ \sec(\theta) = \frac{ 1 }{ \cos(\theta) }$

EXPLANATION:

(i)

$\cos(\theta) = \sqrt{1 - \frac{1}{ {12}^{2} } } \\ = \sqrt{ \frac{ {12}^{2} - 1 }{144} } \\ = \frac{ \sqrt{143} }{12} \\ = \frac{1}{ \cos(\theta) } - \frac{ \sin(\theta) }{ \cos(\theta) } \\ = \frac{1 - \sin(\theta) }{ \cos(\theta) } \\ = \frac{1 - \frac{1}{12} }{ \frac{ \sqrt{143} }{12} } \\ = \frac{ \frac{12 - 1}{12} }{ \frac{ \sqrt{143} }{12} } \\ \sec(\theta) - \tan(\theta) = \frac{11}{ \sqrt{143} }$

(ii)

$= {(\frac{1}{ \cos})^{2} } - {( \frac{ \sin(\theta) }{ \cos(\theta) })}^{2} \\ \frac{1}{ { \cos }^{2} \theta} - \frac{{ \sin}^{2}\theta}{{ \cos}^{2}\theta} \\ = \frac{1 - \frac{1}{144} }{ \frac{143}{144} } \\ = \frac{ \frac{144 - 1}{144} }{ \frac{143}{144} } \\ = \frac{ \frac{143}{144} }{ \frac{143}{144} } \\\frac{1}{ {\cos}^{2}\theta } - { \tan }^{2} \theta = 1$

$\sec(\theta) - \tan(\theta) = \frac{11}{ \sqrt{143} }\\ \frac{1}{ {\cos}^{2}\theta } - { \tan }^{2} \theta = 1$

HERE FROM (ii) we also proved sec²theta -tan²theta=1

Answer 2

First put formula and find the answer and open bracket...