(1+tanA tanB)2+(tanA − tanB)2sec2A sec2B= ___

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\frac{(1 + t a n A t a n B)^{2} + (t a n A - t a n B)^{2}}{{s e c}^{2} A {s e c}^{2} B} =

___

Options:

A . 1

B . -tan A

C . 2

D . cot A

Answer: Option A
:
A

\frac{(1 + tan A tan B)^{2} + (tan A-tan B)^{2}}{{sec}^{2} A {sec}^{2} B}

= \frac{1 + 2 tan A tan B + {tan}^{2} A {tan}^{2} B + {tan}^{2} A - 2 tan A tan B + {tan}^{2} B}{{sec}^{2} A {sec}^{2} B}

= \frac{1 + {tan}^{2} A {tan}^{2} B + {tan}^{2} A + {tan}^{2} B}{{sec}^{2} A {sec}^{2} B}

= \frac{1 + {tan}^{2} A + {tan}^{2} B + {tan}^{2} A {tan}^{2} B}{{sec}^{2} A {sec}^{2} B}

= \frac{1 (1 + {tan}^{2} A) + {tan}^{2} B (1 + {tan}^{2} A)}{{sec}^{2} A {sec}^{2} B}

= \frac{(1 + {tan}^{2} A) (1 + {tan}^{2} B)}{{sec}^{2} A {sec}^{2} B}

= \frac{({sec}^{2} A) ({sec}^{2} B)}{{sec}^{2} A {sec}^{2} B} . \dots (1 + {tan}^{2} θ = {sec}^{2} θ)

= 1

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