Question

$if \: 3 \sin( \beta ) + 4 \cos( \beta ) = 5 \: then \: show \: that \: \sin( \beta ) = 3 \div 5$
solve question no a


no a solve ​

Answer 1

sin∅ = m/n

m = n sin∅ ...(1)

√(n² - m²)tan∅ = m

LHS

= √(n² - n²sin²∅) tan∅

= √[n²(1 - sin²∅)] tan∅

= √n²cos²∅ × tan∅ [• (1 - sin²∅ = cos²∅)]

= n × cos∅ × sin∅/cos∅

= n sin∅

= m ....[From (1)]

Hence Proved ... ✓✓

Answer 2

Answer :-

sin∅ = m/n

m = n sin∅ ...(1)

√(n² - m²)tan∅ = m

LHS

= √(n² - n²sin²∅) tan∅

= √[n²(1 - sin²∅)] tan∅

= √n²cos²∅ × tan∅ [• (1 - sin²∅ = cos²∅)]

= n × cos∅ × sin∅/cos∅

= n sin∅

= m ....[From (1)]

Hence Proved ... ✓✓