1.

If tan 35º = k, then the value oftan 145º – tan 1250/1+tan 145º tan 125​

Answer»

Answer:

145° = 35°+35°+35°+35°+5°

= 4 (35°) +5°

= 4k+5

1250° = 35(35°) +25

= 35k+25

:.tan145° -tan1250°

= 4k+5-(35k+25)

= 4k+5-35k-25

=-31k -20

tan125°=3(35°) +20

= 3k+20

:.tan145° - tan1250°/1+tan145° tan125°

= -31k-20 /(4k+5)(3k+20)



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