Transformation of Trigonometric Formula

Here,
2cos70°.cos20° = cos50°

Taking L.H.S,
= 2cos70°.cos20°
= cos(70° + 20°) + cos(70° - 20°)
= cos90° + cos50°
= 0 + cos50°
∴ cos50° = R.H.S

Here,
$\frac{\text{sin80° + sin10°}}{\text{cos10° - cos80°}}$ = cot35°

Taking L.H.S,
= $\frac{\text{sin80° + sin10°}}{\text{cos10° - cos80°}}$
= $\frac{2sin\left(\frac{\text{80° + 10°}}{2}\right)cos\left(\frac{\text{80° - 10°}}{2}\right)}{2sin\left(\frac{\text{80° + 10°}}{2}\right)sin\left(\frac{\text{80° - 10°}}{2}\right)}$
= $\frac{cos35°}{sin35°}$
∴ cot35° = R.H.S

Here,
$\frac{\text{cosA - cos5A}}{\text{sin5A - sinA}}$ = tan3A

Taking L.H.S,
= $\frac{\text{cosA - cos5A}}{\text{sin5A - sinA}}$
= $\frac{2sin\left(\frac{\text{5A + A}}{2}\right)sin\left(\frac{\text{5A - A}}{2}\right)}{2cos\left(\frac{\text{5A + A}}{2}\right)sin\left(\frac{\text{5A - A}}{2}\right)}$
= $\frac{sin3A}{cos3A}$
∴ tan3A = R.H.S

Here,
$\frac{\text{cos40° - sin30°}}{\text{sin60° - cos50°}}$ = tan50°

Taking L.H.S,
= $\frac{\text{cos40° - cos(90° - 30°)}}{\text{sin60° - sin(90° - 50°)}}$
= $\frac{\text{cos40° - cos60°}}{\text{sin60° - sin40°}}$
= $\frac{2sin\left(\frac{\text{60° + 40°}}{2}\right)sin\left(\frac{\text{60° - 40°}}{2}\right)}{2cos\left(\frac{\text{60° + 40°}}{2}\right)sin\left(\frac{\text{60° - 40°}}{2}\right)}$
= $\frac{sin50°}{cos50°}$
∴ tan50° = R.H.S

Here,
$\frac{\text{cos20° - sin20°}}{\text{cos20° + sin20°}}$ = tan25°

Taking L.H.S,
= $\frac{\text{cos20° - cos(90° - 20°)}}{\text{cos20° + cos(90° - 20°)}}$
= $\frac{\text{cos20° - cos70°}}{\text{cos20° + cos70°}}$
= $\frac{2sin\left(\frac{\text{70° + 20°}}{2}\right)sin\left(\frac{\text{70° - 20°}}{2}\right)}{2cos\left(\frac{\text{20° + 70°}}{2}\right)cos\left(\frac{\text{20° - 70°}}{2}\right)}$
= $\frac{sin45°.sin25°}{cos45°.cos(-25)°}$
= $\frac{sin45°.sin25°}{cos45°.cos25°}$ [cos(-θ) = cosθ]
= $\frac{sin25°}{cos25°}$ [sin45° = cos45° = $\frac{1}{\sqrt2}$]
∴ tan25° = R.H.S

Here,
cos20° + cos140° + cos100° = 0

Taking L.H.S,
= cos20° + cos140° + cos100°
= 2cos$\left(\frac{\text{20° + 140°}}{2}\right)$cos$\left(\frac{\text{20° - 140°}}{2}\right)$ + cos100°
= 2cos80°.cos(-60)° + cos100° [cos(-θ) = cosθ]
= 2cos80°cos60° + cos100°
= $\frac 22$cos80° + cos100°
= 2cos$\left(\frac{\text{80° + 100°}}{2}\right)$cos$\left(\frac{\text{80° - 100°}}{2}\right)$
= 2cos90°.cos(-10)°
∴ 0 = R.H.S

Here,
$\frac 12$(cos2θ - cos8θ) = sin5θ.sin3θ

Taking L.H.S,
= $\frac 12$(cos2θ - cos8θ)
= $\frac 12 \times 2sin\left(\frac{\text{8θ - 2θ}}{2}\right)sin\left(\frac{\text{8θ + 2θ}}{2}\right)$

∴ sin3θ.sin5θ = R.H.S

Here,
2sin50°.sin40° = cos10°

Taking L.H.S,
= 2sin50°.sin40°
= cos(50 - 40°) - cos(50° + 40°)
= cos10° - 0

∴ cos10° = R.H.S

Here,
cos105°.cos15° = -$\frac 14$

Taking L.H.S,
= $\frac 22$ × cos105°.cos15°
= $\frac{\text{cos(105° + 15°) + cos(105° - 15°)}}{2}$
= $\frac{\text{cos120° + cos90°}}{2}$
= $\frac{cos120°}{2}$

∴-$\frac{1}{4}$ = R.H.S

Here,
sin75° - sin105°

Taking L.H.S,
= 2cos$\left(\frac{\text{75° + 105°}}{2}\right)$sin$\left(\frac{\text{75° - 105°}}{2}\right)$
= 2cos90° × sin(-15)°
= 2 × 0 × sin(-15)°
= 0

Here,
$\frac{\text{cos40° - sin40°}}{cos40° + sin40°}$ = tan5°

Taking L.H.S,
= $\frac{\text{cos40° - cos(90° - 40°)}}{\text{cos40° + cos(90° - 40°)}}$
= $\frac{\text{cos40° - cos50°}}{\text{cos40° + cos50°}}$
= $\frac{2sin\left(\frac{\text{50° - 40°}}{2}\right)sin\left(\frac{\text{50° + 40°}}{2}\right)}{2cos\left(\frac{\text{40° + 50°}}{2}\right)cos\left(\frac{\text{40° - 50°}}{2}\right)}$
= $\frac{sin5°.sin45°}{cos45°.cos(-5)°}$
= $\frac{sin5°}{cos5°}$ [cos(-θ) = cosθ, sin45° = cos45° = $\frac{1}{\sqrt2}$]

∴ tan5° = R.H.S

Here,
cos40° + sin40° = $\sqrt 2$cos5°

Taking L.H.S,
= cos40° + cos(90° - 40°)
= cos40° + cos50°
= 2cos$\left(\frac{\text{40 + 50}}{2}\right)$.cos$\left(\frac{\text{40 - 50}}{2}\right)$
= 2cos45°.cos(-5)°
= $\frac{2}{\sqrt2}$cos5° [cos(-θ) cosθ]
= $\frac{\sqrt2 \times \sqrt2}{\sqrt2}$cos5°

∴ $\sqrt2$cos5° = R.H.S

Here,
$\frac{\text{sinA + sinB}}{\text{sinA - sinB}}$ = tan$\frac{\text{A + B}}{2}$.cot$\frac{\text{A - B}}{2}$

Taking L.H.S,
= $\frac{2sin\left(\frac{\text{A + B}}{2}\right)cos\left(\frac{\text{A - B}}{2}\right)}{2sin\left(\frac{\text{A - B}}{2}\right)cos\left(\frac{\text{A + B}}{2}\right)}$

∴ tan$\frac{\text{A + B}}{2}$.cot$\frac{\text{A - B}}{2}$ = R.H.S