Question

cos 13°
sin 13°)
Show that:
(i) tan 10° tan 15° tan 75º tan 80° = 1
(ii) sin 42° sec 48° + cos 42° cosec 48° = 2
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Answer 1

$\small \purple {\sf{ Question !}}$

(i) tan 10° tan 15° tan 75º tan 80° = 1

$\small \blue {\sf{ Answer :- }}$

L.H.S. = tan 10˚ tan 15˚ tan 75˚ tan 80˚

= tan 10˚ tan 15˚ tan (90˚ – 15˚) tan(90˚ – 10˚)

= tan 10˚ tan 15˚ cot 15˚ cot 10˚

1/cot10˚ × 1/cot 15˚ × cot 15˚ × cot 10˚

= 1 = R.H.S.

Hence proved.

Answer 2

Step-by-step explanation:

I) tan 10° tan15° tan75° tan80°

tan10° tan(90-80)° × tan15°tan(90-75)°

tan10°cot10° × tan15°cot15°

tan10° 1/tan 10° × tan15°1/tan15°

1×1 =1

lhs= rhs

ii)sin42°sec(90-48)° + sec(90-42)°cosec48°

sin42°cos42° + sec48°coses48°

sin42°×1/sin 42° + sec 48°×1/sec48°

1 +1 =2

lhs=rhs