Question

y
+
b
= 1.
6. If x = a cos30 and y = b sin’e, prove that (A ý:
7. If (tan 0+ sin 0) = m and (tan 0-sin ) = n, prove that
(m? – n2)2 = 16mn.
8. If (cot 0 + tan 0) = m and (sec 0 - cos 0) = n, prove that
(mn)%3– (mn?)% = 1.
9. If (cosec 0 - sin 0) = a' and (sec 0 - cos 0) = b), prove that
a2b2(a² + b2) = 1.
10. If (2 sin 0+3 cos 0) = 2, prove that (3 sin 0 -2 cos 0) = -3.
11. If (sin 0 + cos 6) = 72 cos 0, show that cot 0 = (y2 +1).
12. If (cos 0 + sin 0) = 12 sin 0, prove that (sin 0 - cos 0) = 12 cos 0.
13. If sec 0 + tan 0 =p, prove that
1
p2-1
(i) sec 0 =
(ii) tan =
(iii) sin e
2
2
p2+1
2(p+1
(p-1​

Answer 1

Answer:anθ+sinθ=m and tanθ-sinθ=n

∴, m²-n²

=(m+n)(m-n)

=(tanθ+sinθ+tanθ-sinθ)(tanθ+sinθ-tanθ+sinθ)

=(2tanθ)(2sinθ)

=4tanθsinθ

4√mn

=4√(tanθ+sinθ)(tanθ-sinθ)

=4√(tan²θ-sin²θ)

=4√{(sin²θ/cos²θ)-sin²θ}

=4√sin²θ{(1/cos²θ)-1}

=4sinθ√{(1-cos²θ)/cos²θ}

=4sinθ√(sin²θ/cos²θ)

=4sinθ√tan²θ

=4sinθtanθ

∴, LHS=RHS (Proved)

Tan θ + sinθ = m Squaring on both sides& Tan θ - Sin θ = n Squaring on both sides we get

prove that (m2 - n2)²=16mn

Tan2 θ + sin2 θ + 2 Tan θ sinθ = m2 ---------(1)

Tan2 θ + sin2 θ - 2 Tan θ sinθ = n2 --------(2)

Substract (2) from (1) we get

m2 - n2 = 4 Tan θ sinθ = 4 √( Tan2 θ  sin2 θ)

 = 4 √( Sin2 θ /Cos2 θ ( 1 - Cos2 θ) )

 = 4 √( Tan2 θ - sin2 θ)

= 4 √( Tan θ + Sin θ)(Tan θ - Sin θ )

m2 - n2 = 4√mn ( on squaring both sides)

(m2-n2)²=16mn