Question

If cos (x-y) cos(z-t) = cos(x+y) cos (z+t), then tanx tany + tanz tant is equal to

• A. 1
• B. -1
• C. 2
• D. 0

Answer 1

The correct option is D

0

We will first bring the terms which has x and y on one side and z and t on the other side. After that we will apply componendodividendo (We need tanx, tany, tanz and tant. After solving similar problems, it becomes obvious that componendodividendo can lead to terms which has tan)

(After getting the similar temrs on one side we can expand cos terms and then divide by proper ratios (cosxcosy and coszcost) to get tanx, tany, tanz

$\frac{cos(x-y)}{cos(x+y)}$ = $\frac{cos(z+t)}{cos(z-t)}$

$\Rightarrow$ $\frac{cos(x-y)+cos(x+y)}{cos(x-y)-cos(x+y)}$ = $\frac{cos(z+t)+cos(z-t)}{cos(z+t)-cos(z-t)}$

$\Rightarrow$ $\frac{2cosxcosy}{2sinxsiny}$ = $\frac{2coszcost}{-2sinzsint}$

$\Rightarrow$ tanx tany = - tanz tant

$\Rightarrow$ tanx tany + tanz tant = 0

key steps/concepts: (1) cosA + cosB = 2cos$\frac{A+B}{2}$ cos $\frac{(A-B)}{2}$

(2) cosA - cosB = 2sin$\frac{(A+B)}{2}$ sin $\frac{(B-A)}{2}$

(3) Applying componendo-dividendo

and tan t