View source for Activity-trigonometry problems


='''Problem-1'''=
'''prove that''' <math>\frac{1-\tan^2 A}{1+\tan^2 A}=1-2\sin^2 A</math>

==Interpretation of problems==
#It is to prove  the problem  based on trigonometric identities
#the function of one trigonometric ratio is relates to other
==Concept development==
Develop the skill of proving problem based trigonometric identity  
==Skill development==
Problem solving
==Pre Knowledge require==
# Idea about trignometric ratios
# Idea about trignometric identities
==Methos Of Solutions==
=== '''Generalisation By Verification'''===
When A=60°
LHS=<math>\frac{1-\tan^2 60°}{1+\tan^2 60°}</math> <br>=<math>\frac{1-{(\sqrt{3})}^2 }{1+{(\sqrt{3})}^2 }</math><br>=<math>\frac{1-3}{1+3}</math><br>=<math>\frac{-2}{4}</math><br>=<math>\frac{-1}{2}</math>-----(1)<br>RHS=<math>1-2\sin^2 A</math><br>=<math>1-2\sin^260° </math><br><math>1-2{(\frac{\sqrt{3}}{2})}^2</math><br>=<math>1-2(\frac{3}{4})</math><br>=<math>\frac{4-2(3)}{4}</math><br>=<math>\frac{-1}{2}</math>------(2)<br>from eqn1 & eqn2<br><math>\frac{1-\tan^2 60°}{1+\tan^2 60°}</math>=<math>1-2\sin^260° </math><br>'''By Generalisation'''<br> '''<math>\frac{1-\tan^2 A}{1+\tan^2 A}=1-2\sin^2 A</math>'''

==='''By Deductive Proof'''===
LHS=<math>\frac{1-\tan^2 A}{1+\tan^2 A}</math><br>=<math>\frac{1-\frac{\sin^2A}{\cos^2A}}{\sec^2A}</math><br>
=<math>\frac{[\cos^2A-\sin^2A]}{cos^2A}\cos^2A</math><br>
=<math>1-\sin^2A-\sin^2A</math><br>=<math>1-2\sin^2A</math>=RHS

[[Category:Trigonometry]]