InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
If \(\sin \theta = \sqrt {\frac{{{x^2} - {y^2}}}{{{x^2}+ {y^2}}}}\), then the value of \(\sqrt {({{\sec }^2}\theta {\rm} + {\rm}{{\tan }^2}\theta})\) is (where 0° < θ < 90°) :1). \(\frac{{{x^2}}}{{x\sqrt {{y^2} - {x^2}} }}\)2). 13). x/y4). y/x |
|
Answer» $(\sin \theta = \sqrt {\frac{{{x^2} - {y^2}}}{{{x^2}+ {y^2}}}} )$ We know that, sin θ = Perpendicular/HYPOTENUSE where, Perpendicular $(= {\rm}\sqrt {{x^2} - {y^2}} )$ and Hypotenuse $(= {\rm}\sqrt {{x^2} + {y^2}} )$ We know according to Pythagoras THEOREM, (Hypotenuse)2 = (Perpendicular)2 + (BASE)2 ⇒ x2 + y2 = x2 – y2 + (Base)2 ⇒ Base $(= {\rm}\sqrt {{x^2}+ {y^2} - {x^2} + {y^2}} = \sqrt 2 y)$ SEC θ = Hypotenuse/Base $(= {\rm}\left( {\sqrt {{x^2}+{y^2}}} \right)/\sqrt 2 y)$ tan θ = Perpendicular/Base $(= {\rm}\left( {\sqrt {{x^2} - {y^2}}} \right)/\sqrt 2 y)$ $(\sqrt {({{\sec }^2}\theta {\rm} + {\rm}{{\tan }^2}\theta })= \sqrt {\frac{{{x^2}+ {y^2}}}{{2{y^2}}}+ \frac{{{x^2} - {y^2}}}{{2{y^2}}}} = \frac{x}{y})$ |
|