Question

V)
tan(Q + B) = a + b 47 tan(Q - B) = a - b 03 065
(palis (3 ataną - btan B = a 2 – b2​

Answer 1

Step-by-step explanation:

(1+tan15°)1−tan15°)(1+tan15°)1−tan15°)

=

we know that

Tan(x2=1−cosxsinxTan(x2=1−cosxsinx

=>

tan15°=1−cos30°sin30°tan15°=1−cos30°sin30°

Therefore :

(1+1−cos30°sin30°)1−1−cos30°sin30°)(1+1−cos30°sin30°)1−1−cos30°sin30°)

=> =

(sin30°+1−cos30°)(sin30°−1+cos30°)(sin30°+1−cos30°)(sin30°−1+cos30°)

But we know that

cos30°=3√2cos30°=32

and

sin30°=12sin30°=12

=>

(sin30°+1−cos30°)(sin30°−1+cos30°)(sin30°+1−cos30°)(sin30°−1+cos30°)

=

(12+1−3√2)(12−1+3√2)(12+1−32)(12−1+32)

=

(1+2−3√)(1−2+3√)(1+2−3)(1−2+3)

=

(3−3√)(−1+3√)(3−3)(−1+3)

=

(3−3√)(1+3√)(−1+3√)(1+3√)(3−3)(1+3)(−1+3)(1+3)

=

(3+3.3√−3√−3)(3−1)(3+3.3−3−3)(3−1)

=

(2.3√2(2.32

=

3–√3