`sin^2 beta + cos^2 beta` = ??

The value of sin^2 alpha cos^2 beta + cos^2 alpha sin^2 beta + sin^2 alpha sin^2 beta +cos^2 alpha cos^2 beta is

cos^(2)theta+sin^(2)theta cos2 beta=cos^(2)beta+sin^(2)beta cos2 theta

Prove that: cos^(2)theta+sin^(2)theta cos2 beta=cos^(2)beta+sin^(2)beta cos2 theta

If 2tan alpha=3tan beta, then tan (alpha-beta)=(sin2 beta)/(5-cos2 beta)(b)(cos2 beta)/(5-cos2 beta)(c)(sin2 beta)/(5+cos2 beta)(d) none of these

If tan(alpha-beta)=(sin2 beta)/(3-cos2 beta), then ( i )tan(alpha)=2tan(beta)( ii) (1)/(2)tan(alpha)=tan(beta)( (ii) 2tan(alpha)=3tan(beta)( iv) 3tan(alpha)=2tan(beta)

If tan(alpha-beta)=(sin2 beta)/(3-cos2 beta) then (a) tan alpha=2tan beta(b)tan beta=2tan alpha(c)2tan alpha=3tan beta(d)3tan alpha=2tan beta

If A=[(cos^(2)alpha, cos alpha sin alpha),(cos alpha sin alpha, sin^(2)alpha)] and B=[(cos^(2)betas,cos beta sin beta),(cos beta sin beta, sin^(2) beta)] are two matrices such that the product AB is null matrix, then alpha-beta is

Evaluate: quad =|cos alpha cos beta cos alpha sin beta-sin alpha-sin beta cos beta0sin alpha cos beta sin alpha sin beta cos alpha|