Question

Prove :-
Tan48° Tan23° Tan42° Tan67° = 1​

Answer 1

Answer:

$we \: know \\ that  \: tan \: (90−θ) \: = \: cotθ  \: \\ and  \\ \: cos(90−θ)=sinθ \\ </p><p></p><p></p><p>(i) tan \: (90−42) \: otan \\ \\ (90−67) \: otan \: 42 \: otan \: 67 \: \\ o</p><p></p><p>=cot \: 42 \: o \: cot \: 67 \: \\ o \: tan \: 42 \: o \: tan \: 67 \: o \\ \\ </p><p></p><p>= \: (1) \: (1) \: =1 \\ </p><p></p><p></p><p>(ii\: cos(96−52) \\ o \:cos(90−38) \: o−sin \: 38 \: o \: sin \: 52 \: o \\ </p><p></p><p>=sin \: 52 \: osin \: 38 \: o \: −sin \: 38 \: osin \: 52 \: o</p><p> \\ </p><p>=0 \\ \\ </p><p></p><p>Hence \: Proved.</p><p></p><p>$

Answer 2

Solution:

=Tan48° Tan23° Tan42° Tan67° = 1

=Cot(90° - 48°) Cot(90° - 23°) Tan42° Tan67°

=Cot42° Cot67° Tan42° Tan67°

$\frac{1}{tan42°} \times \frac{1}{tan67°} \times tan42 ° \times tan67°$

= 1 = RHS

Hence, proved!