[tex]\frak{\pmb{\underline\red{Question}}}: -[/tex]
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If A and B are complementary angles then _________
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(a) sinA = cosecB
(b) tanA = tanB
(c) cosA = secB
(d) cosecA = secB
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[tex]\frak{\pmb{\underline\red{Explanation \: needed!!}}}[/tex]
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Verified answer
Answer:
A and B are complementary angles i.e A = 90°, B = 90°
sinA = sin90° = 1
cosecB = cosec90° = 1
Therefore, sinA = cosecB
Ans. a) sinA = cosecB
Hope it helps.
❍ Solution :-
Given that A and B are complementary angles.
A + B = 90°
A + B = 90°
A = 90° - B
On putting cosec on both sides,
cosec A = cosec (90° - B)
cosec A = sec B
[cosec (90° - θ) = sec θ]
The correct option is (d) cosec A = sec B
Trigonometric Ratios :-
[tex]\begin{gathered}\begin{array}{ | c|c|c|c|c|c |}\hline \rm\angle\:A& \: \: \: 0\degree&30\degree&45\degree&60\degree&90\degree\\ \hline \rm\sin \: A&0& \dfrac{ 1}{2}&\dfrac{1}{\sqrt{2}}&\dfrac{\sqrt{3}}{2}&1\\ \hline \rm\cos \:A&1& \dfrac{ \sqrt{3} }{2} & \dfrac{1}{ \sqrt{2}} & \dfrac{1}{2}&0 \\ \hline \rm \tan \: A&0& \dfrac{1}{ \sqrt{3} }&1& \sqrt{3}& \rm{ \infty } \\ \hline \rm\cosec \: A& \infty & 2& \sqrt{2} & \dfrac{2}{\sqrt{3} }&1 \\ \hline \rm\sec \: A&1& \dfrac{2}{ \sqrt{3} }& \sqrt{2}&2 & \infty \\ \hline \rm \cot \: A& \infty & \sqrt{3} &1& \dfrac{1}{ \sqrt{3} }&0 \\ \hline \end{array}\end{gathered}[/tex]