Difference between revisions of "Activity-trigonometry problems"
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='''Problem-1'''= | ='''Problem-1'''= | ||
| − | '''prove that''' <math>\frac{1-\tan^2 A}{1+\tan^2 A}=1-\sin^2 A</math> | + | '''prove that''' <math>\frac{1-\tan^2 A}{1+\tan^2 A}=1-2\sin^2 A</math> |
==Interpretation of problems== | ==Interpretation of problems== | ||
| Line 16: | Line 16: | ||
# '''Generalisation By Verification''' | # '''Generalisation By Verification''' | ||
When A=60° | 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) | + | 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>''' | ||
Revision as of 23:53, 31 July 2014
Problem-1
prove that
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
- Generalisation By Verification
When A=60°
LHS=Failed to parse (syntax error): {\displaystyle \frac{1-\tan^2 60°}{1+\tan^2 60°}}
=
=
=
=-----(1)
RHS=
=Failed to parse (syntax error): {\displaystyle 1-2\sin^260° }
=
=
=------(2)
from eqn1 & eqn2
Failed to parse (syntax error): {\displaystyle \frac{1-\tan^2 60°}{1+\tan^2 60°}}
=Failed to parse (syntax error): {\displaystyle 1-2\sin^260° }
By Generalisation
