If 0 < α < π 16 and ( 1 + tan ⁡ α ) 1 + tan ⁡ 4 α = 2 , then the value of α is equal to

Step-by-step solution

We have 1 + tan &#8289; &#945; + tan &#8289; 4 &#945; + tan &#8289; &#945; tan &#8289; 4 &#945; = 2 tan &#8289; &#945; + tan &#8289; 4 &#945; = 1 - tan &#8289; &#945; tan &#8289; 4 &#945; tan &#8289; &#945; + tan &#8289; 4 &#945; 1 - tan &#8289; &#945; tan &#8289; 4 &#945; = 1 tan &#8289; &#945; + 4 &#945; = 1 5 &#945; = &#960; 4 s i n c e , &#160; 0 &#60; &#945; &#60; &#960; 16 Hence, &#945; = &#960; 20