Question

$\frak{\pmb{\underline\red{Question}}}: -$

If A and B are complementary angles then _________

(a) sinA = cosecB

(b) tanA = tanB

(c) cosA = secB

(d) cosecA = secB

$\frak{\pmb{\underline\red{Explanation \: needed!!}}}$
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Answer 1

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.

Answer 2

❍ 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 :-

$\begin{gathered}\begin{array}{ | c|c|c|c|c|c |}\hline \rm\angle\:A& \: \: \: 0\degree&30\degree&45\degree&60\degree&90\degree\\ \hline \rm\sin \: A&0& \dfrac{ 1}{2}&\dfrac{1}{\sqrt{2}}&\dfrac{\sqrt{3}}{2}&1\\ \hline \rm\cos \:A&1& \dfrac{ \sqrt{3} }{2} & \dfrac{1}{ \sqrt{2}} & \dfrac{1}{2}&0 \\ \hline \rm \tan \: A&0& \dfrac{1}{ \sqrt{3} }&1& \sqrt{3}& \rm{ \infty } \\ \hline \rm\cosec \: A& \infty & 2& \sqrt{2} & \dfrac{2}{\sqrt{3} }&1 \\ \hline \rm\sec \: A&1& \dfrac{2}{ \sqrt{3} }& \sqrt{2}&2 & \infty \\ \hline \rm \cot \: A& \infty & \sqrt{3} &1& \dfrac{1}{ \sqrt{3} }&0 \\ \hline \end{array}\end{gathered}$