The value of $\frac{sin23°cos67°tan45°+cos23°sin67°cos45°}{2sin45°cos45°}$ is :
Answer & explanation
Correct answer: option 3
$\frac{sin23°cos67°tan45°+cos23°sin67°cot45°}{2sin45°cos45°}$
We know ,
sinA . cosB + cosA . sinB = sin (A + B)
& tan 45° = 1 , cot 45° = 1
= \(\frac{ sin23°cos67°tan45°+cos23°sin67°cos45° }{2sin45°cos45° }\)
= \(\frac{ sin23°cos67°+cos23°sin67° }{2sin45°cos45° }\)
= \(\frac{ sin ( 23°+67°) }{2 × 1/√2 × 1/√2 }\)
= 1