Question

what is the value of 'tan7 1/2' degree ? or tan7.5 value​

Answer 1

Answer:

Step-by-step explanation:

7½° lies in the first quadrant.

Therefore, both sin 7½° and cos 7½° is positive.

For all values of the angle A we know that, sin (α - β) = sin α cos β - cos α sin β.

Therefore, sin 15° = sin (45° - 30°)

                        = 1√2∙√32 - 1√2∙12

                        = √32√2 - 12√2

                        = √3−12√2

Again, for all values of the angle A we know that, cos (α - β) = cos α cos β + sin α sin β.

Therefore, cos 15° = cos (45° - 30°)

cos 15° = cos 45° cos 30° + sin 45° sin 30°

          = 1√2∙√32 + 1√2∙12

          = √32√2 + 12√2

          = √3+12√2

Now, tan 7½° = sin7½°cos7½°

                   = 2sin27½°2cos7½°sin7½°

                   = 1−cos15°sin15°

                   = 1−√3+12√2√3−12√2

                   = 2√2−√3−1√3−1

                   = (2√2−√3−1)(√3+1)(√3−1)(√3+1)

                   = 2√6−3−√3+2√2−√3−112

                   = √6 - √3 + √2 - 2

Therefore, tan 7½° = √6 - √3 + √2 - 2

THANK U..

Answer 2

tan15o=tan(60–45)o=tan60o−tan45o1+tan60o.tan45otan⁡15o=tan⁡(60–45)o=tan⁡60o−tan⁡45o1+tan⁡60o.tan⁡45o

⟹tan150=3–√−13–√+1⟹tan⁡150=3−13+1

Let θ=7.50⟹2θ=15oθ=7.50⟹2θ=15o

Lett=tanθ=tan7.5oLett=tan⁡θ=tan⁡7.5o

⟹tan2θ=tan15o=3–√−13–√+1⟹tan⁡2θ=tan⁡15o=3−13+1

tan15o=tan2θ=2tanθ1−tan2θ=3–√−13–√+1tan⁡15o=tan⁡2θ=2tan⁡θ1−tan2⁡θ=3−13+1

⟹2t1−t2=