Question 139
$$\angle Y$$ is the right angle of the triangle $$XYZ$$. If $$XY=2\sqrt{6}$$ cm and $$XZ-YZ=2$$ cm, then the value of $$(secX + tanX)$$ is:
Solution
Given : $$XY=2\sqrt{6}$$ cm and $$XZ-YZ=2$$
To find : $$(secX + tanX)$$ = ?
Solution : In $$\triangle$$ XYZ,
=> $$(XY)^2=(XZ)^2-(YZ)^2$$
=> $$(2\sqrt6)^2=(XZ-YZ)(XZ+YZ)$$
=> $$(2)(XZ+YZ)=24$$
=> $$(XZ+YZ)=\frac{24}{2}=12$$ -------------(i)
$$\therefore$$Â $$(secX + tanX)$$
= $$(\frac{XZ}{XY})+(\frac{YZ}{XY})=\frac{(XZ+YZ)}{XY}$$
= $$\frac{12}{2\sqrt6}=\sqrt6$$
=> Ans - (D)
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