Question

If sin theta=3/5 prove (tan theta+sec theta)^2=4​

Answer 1

$\large{ \rm \bull \underline{ \underline{Given:-}}}$

• $\sin \theta = \dfrac{3}{5}$

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$\large{ \rm \bull \underline{ \underline{To \: Prove :-}}}$

• $( \tan \theta + \sec \theta )^{2} = 4$

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$\large{ \rm \bull \underline{ \underline{Proof :-}}}$

Substitute the given values sin θ

$\qquad{ ━━━━━━━━━━━━━━━━━━━━━━━━━━━}$

• ➻$\rm \sin \theta = \dfrac{3}{5} = \dfrac{Opposite \: side}{Hypotenuse}$
• ➻$\rm \tan \theta = \dfrac{3}{?} = \dfrac{Opposite \: side}{Adjacent \: side}$
• ➻$\rm \sec \theta = \dfrac{5}{?} = \dfrac{Hypotenuse}{Adjacent \: side}$

$\qquad{ ━━━━━━━━━━━━━━━━━━━━━━━━━━━}$

AB = 3

AC = 5

BC = ?

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To find the missing side we need to use Pythagoras theorem.

$\dag \underline{ \boxed{\rm{Hypotenuse ^{2} = Opposite ^{2} +Adjacent ^{2} }}} \dag$

$\dag \underline{ \boxed{\rm Missing \:Side² = Hypotenuse²- Adjacent\: Side² }}\dag$

$: \longmapsto \rm BC^{2} = {5}^{2} - {3}^{2}$

$: \longmapsto \rm BC= \sqrt{ {5}^{2} - {3}^{2} }$

$: \longmapsto \rm BC = \sqrt{ 25- 9 }$

$: \longmapsto \rm BC = \sqrt{ 16 }$

$: \longmapsto \rm BC = 4$

So the other side is = 4

$: \longmapsto \rm ( \tan \theta + \sec \theta )^{2} = 4$

$: \longmapsto \rm \bigg\lgroup \tan \theta = \dfrac{3}{4} + \sec \theta= \dfrac{5}{4} \bigg\rgroup ^{2}$

$: \longmapsto \rm \bigg\lgroup \dfrac{3}{4} + \dfrac{5}{4} \bigg\rgroup ^{2}$

$: \longmapsto \rm \bigg\lgroup \dfrac{8}{4} \bigg\rgroup ^{2}$

$: \longmapsto \rm \bigg\lgroup \cancel{ \dfrac{8}{4} } \bigg\rgroup ^{2}$

$: \longmapsto \rm \big\lgroup 2 \big\rgroup ^{2}$

$: \longmapsto \rm 4$

Hence Proved

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Additional Information:-

Ratios:-

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

Identities:-

• sin(90° - A) = cos A
• cos(90° - A) = sin A
• tan(90° - A) = cot A
• cot(90° - A) = tan A
• sec(90° - A) = cosec A
• cosec(90° - A) = sec A
• sin² + cos² = 1
• sec² - tan² = 1
• 1 + tan² = sec²
• cosec² = 1 + cot²

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