=08 पथ + 08 पथ - 0४ पथ « 08 पथ - 01 पथ) §p+ 2

1.	=08 पथ + 08 पथ - 0४ पथ « 08 पथ - 01 पथ) §p+ 2 s g 2
Answer» Sin 18°/cos 72° + root(3) [tan10 tan 30 tan 40 tan 50 tan 80] = Sin18/cos(90-18) + root(3) (cot(90-10). tan 30 cot(90-40). tan50. tan 80)= sin18/sin18 + root(3)(cot 80. tan 30. cot50. tan50. tan 80)= 1 + root(3)(tan30)= 1 + root(3)[1/root(3)]= 1 + 1= 2

=08 पथ + 08 पथ - 0४ पथ « 08 पथ - 01 पथ) §p+ 2 s g 2

Answer»

Sin 18°/cos 72° + root(3) [tan10 tan 30 tan 40 tan 50 tan 80]

= Sin18/cos(90-18) + root(3) (cot(90-10). tan 30 cot(90-40). tan50. tan 80)= sin18/sin18 + root(3)(cot 80. tan 30. cot50. tan50. tan 80)= 1 + root(3)(tan30)= 1 + root(3)[1/root(3)]= 1 + 1= 2

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=08 पथ + 08 पथ - 0४ पथ « 08 पथ - 01 पथ) §p+ 2 s g 2

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