Question

Tan pi/12 Tan pi/16 Tan 5pi/12 Tan 7pi/16

Answer 1

it is given that, tan(π/12) . tan(π/16) . tan(5π/12) . tan(7π/16).

we know, tan(π/2 - α) = cotα ......(1)

so, tan(π/12) = tan(π/2 - 5π/12)

= cot(5π/12) [ from equation (1) ]

so, tan(π/12) = cot(5π/12)

again, tan(π/16) = tan(π/2 - 7π/16)

= cot(7π/16) [ from equation (1) ]

so, tan(π/16) = cot(7π/16)

now, tan(π/12) . tan(π/16) . tan(5π/12) . tan(7π/16)

= cot(5π/12). cot(7π/16) . tan(5π/12). tan(7π/16)

= {tan(5π/12). cot(5π/12)}.{tan(7π/16).cot(7π/16)}

we know, tanα.cotα = 1

so, tan(5π/12).cot(5π/12) = 1

and tan(7π/12).cot(7π/12) = 1

hence, {tan(5π/12). cot(5π/12)}.{tan(7π/16).cot(7π/16)} = 1 . 1 = 1 [answer]

Answer 2

Answer:

t is given that, tan(π/12) . tan(π/16) . tan(5π/12) . tan(7π/16).

we know, tan(π/2 - α) = cotα ......(1)

so, tan(π/12) = tan(π/2 - 5π/12)

= cot(5π/12) [ from equation (1) ]

so, tan(π/12) = cot(5π/12)

again, tan(π/16) = tan(π/2 - 7π/16)

= cot(7π/16) [ from equation (1) ]

so, tan(π/16) = cot(7π/16)

now, tan(π/12) . tan(π/16) . tan(5π/12) . tan(7π/16)

= cot(5π/12). cot(7π/16) . tan(5π/12). tan(7π/16)

= {tan(5π/12). cot(5π/12)}.{tan(7π/16).cot(7π/16)}

we know, tanα.cotα = 1

so, tan(5π/12).cot(5π/12) = 1

and tan(7π/12).cot(7π/12) = 1

hence, {tan(5π/12). cot(5π/12)}.{tan(7π/16).cot(7π/16)} = 1 . 1 = 1 [answer]

Step-by-step explanation: