| If \[\tan A=n\tan B\] and \[\sin A=m\,\sin B,\] then the value of \[{{\cos }^{2}}A\] is [SSC (CGL) Mains 2015] |
A) \[\frac{{{m}^{2}}-1}{{{n}^{2}}+1}\]
B) \[\frac{{{m}^{2}}+1}{{{n}^{2}}+1}\]
C) \[\frac{{{m}^{2}}-1}{{{n}^{2}}-1}\]
D) \[\frac{{{m}^{2}}+1}{{{n}^{2}}-1}\]
Correct Answer: C
Solution :
| Given, \[\tan A=n\,\,\tan B\] |
| \[\therefore \] \[\cot B=n\cot A\] ... (i) |
| and \[\sin A=m\,\sin B\] |
| \[\Rightarrow \] \[\text{cosec}\,B=m\,\text{cosec}\,A\] ... (ii) |
| On squaring both sides of Eqs. (i) and (ii) and then subtracting Eq. (i) from Eq. (ii), we get |
| \[\text{cose}{{\text{c}}^{2}}B-{{\cot }^{2}}B\,\,=\,\,{{m}^{2}}\text{cose}{{\text{c}}^{2}}A-{{n}^{2}}{{\cot }^{2}}A\] |
| \[\Rightarrow \] \[\frac{{{m}^{2}}-{{n}^{2}}{{\cos }^{2}}A}{{{\sin }^{2}}A}=1\] |
| \[\Rightarrow \] \[{{m}^{2}}-{{n}^{2}}{{\cos }^{2}}A=1-{{\cos }^{2}}A\] |
| \[\therefore \] \[{{\cos }^{2}}A=\frac{{{m}^{2}}-1}{{{n}^{2}}-1}\] |
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