prove that (1+tan^2 theta ) cos theta cos(90-theta)=cot (90-theta) help me please help me please
Answers
Answer:
the present age of mother be M years
present age of daughter be D years
The sum of ages of mother and her daughter is 60 years.
=> \sf{M\:+\:D\:=\:60}M+D=60
=> \sf{M\:=\:60\:-\:D}M=60−D ....(1)
12 years ago, the mother was eight times as old as her daughter.
12 years ago -
Age of mother = (M - 12) years
Age of daughter = (D - 12) years
According to question,
=> \sf{M\:-\:12\:=\:8(D\:-\:12)}M−12=8(D−12)
=> \sf{M\:-\:12\:=\:8D\:-\:96}M−12=8D−96
=> \sf{60\:-\:D\:-\:12\:=\:8D\:-\:96}60−D−12=8D−96 [From (1)]
=> \sf{48\:-\:D\:=\:8D\:-\:96}48−D=8D−96
=> \sf{48\:+\:96\:=\:8D\:+\:D}48+96=8D+D
=> \sf{144\:=\:9D}144=9D
=> \sf{\frac{144}{9}\:=\:D}
9
144
=D
=> \sf{16\:=\:D}16=D
=> \boxed{\sf{D\:=\:16}}
D=16
\therefore∴ Present age of daughter is 16 years
Substitute value of D = 16 in (1)
=> \sf{M\:=\:60\:-16}M=60−16
=> \boxed{\sf{M\:=\:44}}
M=44
\therefore∴ Present age of mother is 44 years
____________________
Verification :-
From above calculation, we have -
present age of mother = M = 44 years
present age of daughter = D = 16 years
Substitute value of M & D in (1)
→ 44 = 60 - 16
→ 44 = 44
Answer:
tan theta=tan theta
Step-by-step explanation:
sec theta.sec theta.1/sec theta.sin theta=tan theta
sec theta.sin theta=tan theta
1/cos theta.sin theta =tan theta
sin theta/cos theta =tan theta
tan theta=tan theta
Hence Proved
-----------:-thanks