Question

prove this trigonometric equation​

Answer 1

Answer:

check it out... this is done correctly for 3 marks question

Answer 2

$\large\underline{\sf{Given- }}$

$\rm :\longmapsto\: {tan}^{2} \theta \: = {cos}^{2}\phi \: - {sin}^{2}\phi \:$

$\large\underline{\sf{To\:Prove - }}$

$\rm :\longmapsto\: {tan}^{2}\phi \: = {cos}^{2}\theta \: - {sin}^{2}\theta \:$

$\begin{gathered}\Large{\sf{{\underline{Formula\:Used - }}}}  \end{gathered}$

$\boxed{ \bf{ \: 1 + {tan}^{2}x = {sec}^{2}x}}$

$\boxed{ \bf{ \: 1 - {sin}^{2}x = {cos}^{2}x}}$

$\boxed{ \bf{ \: cosx = \dfrac{1}{secx}}}$

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\: {tan}^{2} \theta \: = {cos}^{2}\phi \: - {sin}^{2}\phi \:$

• On adding '1' on both sides, we get

$\rm :\longmapsto\: 1 + {tan}^{2} \theta \: = 1 + {cos}^{2}\phi \: - {sin}^{2}\phi \:$

$\rm :\longmapsto\: {sec}^{2} \theta \: = {cos}^{2}\phi \: + (1 - {sin}^{2}\phi \:)$

$\rm :\longmapsto\: {sec}^{2} \theta \: = {cos}^{2}\phi \: + {cos}^{2}\phi \:$

$\rm :\longmapsto\: {sec}^{2} \theta \: = 2{cos}^{2}\phi \:$

$\rm :\implies\:\dfrac{1}{ {cos}^{2}\theta \: } = \dfrac{2}{ {sec}^{2} \phi \:}$

$\rm :\longmapsto\: {sec}^{2}\phi \: = 2 {cos}^{2}\theta \:$

$\rm :\longmapsto\:1 + {tan}^{2}\phi \: = {cos}^{2}\theta \: + {cos}^{2}\theta \:$

$\rm :\longmapsto\: \cancel1 + {tan}^{2}\phi \: = {cos}^{2}\theta \: + \cancel1 - {sin}^{2}\theta \:$

$\rm :\longmapsto\: {tan}^{2}\phi \: = {cos}^{2}\theta \: - {sin}^{2}\theta \:$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

Additional Information :-

$\boxed{ \bf{ \: {sin}^{2}x + {cos}^{2} x = 1}}$

$\boxed{ \bf{ \: {cosec}^{2} x - {cot}^{2} x = 1}}$

$\boxed{ \bf{ \: {sec}^{2} x - {tan}^{2} x = 1}}$

$\boxed{ \bf{ \: cosecx = \dfrac{1}{sinx} }}$

$\boxed{ \bf{ \: tanx = \dfrac{1}{cotx} = \dfrac{sinx}{cosx} }}$