In the given figure, D is the mid-point of BC, then the value of coty°cotx° is
2
14
13
12
Explanation for the correct option:
Given: D is the mid-point of BC⇒BD=CD
cotx°=ACCD
coty°=ACBCcoty°=ACBD+CDcoty°=ACCD+CD∵BD=CDcoty°=AC2×CD
coty°cotx°=AC2×CDACCDcoty°cotx°=12
Therefore, coty°cotx°=12
Hence, option (D) is the correct option.
In a right ΔABC right-angled at C, if D is the mid-point of BC, prove that BC2=4(AD2−AC2).